代码随想录第二十一天

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复习

深度优先遍历

递归
以中序为例

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class Solution {
    List<Integer> result = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if (root == null) return result;
        inorder(root);
        return result;
    }

    private void inorder(TreeNode node){
        if (node == null) return;
        inorder(node.left);
        result.add(node.val);
        inorder(node.right);
    }
}

迭代
前序

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class Solution {

    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        Stack<TreeNode> stack = new Stack<>();

        stack.push(root);

        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            result.add(node.val);
            if (node.right != null) stack.push(node.right);
            if (node.left != null) stack.push(node.left);
        }  

        return result;
    }
}

中序

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class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;

        while(cur != null || !stack.isEmpty()){
            if (cur != null){
                stack.push(cur);
                cur = cur.left;
            } else {
                cur = stack.pop();
                result.add(cur.val);
                cur = cur.right;
            }
        }

        return result;
    }
}

后序

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class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            result.add(node.val);
            if (node.left != null) stack.push(node.left);
            if (node.right != null) stack.push(node.right);
        }

        Collections.reverse(result);
        return result;
    }
}

统一迭代
以中序为例

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class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            if (node != null){
                if (node.right != null) stack.push(node.right);
                stack.push(node);
                stack.push(null);
                if (node.left != null) stack.push(node.left);
            } else {
                result.add(stack.pop().val);
            }
        }

        return result;
    }
}

DFS

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) return result;
        dfs(root, 0);
        return result;
    }

    private void dfs(TreeNode node, int depth){
        if (node == null) return;
        depth++;

        if (depth > result.size()){
            result.add(new ArrayList<Integer>());
        }

        result.get(depth - 1).add(node.val);

        if (node.left != null) dfs(node.left, depth);
        if (node.right != null) dfs(node.right, depth);
    }
}

广度优先遍历

BFS

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class Solution {
    
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()){
            int length = queue.size();
            List<Integer> itemList = new ArrayList<>();
            while(length-- > 0){
                TreeNode node = queue.poll();
                itemList.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            result.add(itemList);
        }

        return result;
    }
}

226 翻转二叉树

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class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return root;
        swapChildren(root);
        invertTree(root.left);
        invertTree(root.right);

        return root;
    }

    private void swapChildren(TreeNode node){
        TreeNode temp = node.left;
        node.left = node.right;
        node.right = temp;
    }
}

101 对称二叉树

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class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return compare(root.left, root.right);
    }
    private boolean compare(TreeNode left, TreeNode right){
        if (left == null && right == null) return true;
        if (left == null || right == null || left.val != right.val) return false;
        boolean compareOutside = compare(left.left, right.right);
        boolean compareInside = compare(left.right, right.left);

        return compareInside && compareOutside;
    }
}

104 二叉树最大深度

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class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) return 0;
        return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
    }
}

111 二叉树最小深度

递归

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class Solution {
    int mindepth = Integer.MAX_VALUE;
    public int minDepth(TreeNode root) {
        if (root == null) return 0;
        dfs(root,0);

        return mindepth;
    }

    private void dfs(TreeNode root, int depth){
        if (root == null) return;
        depth++;
        if (root.left == null && root.right == null){
            if (mindepth > depth){
                mindepth = Math.min(depth, mindepth);
                return;
            }
        }
        
        if (root.left != null){
            dfs(root.left, depth);
        }
        if (root.right != null){
            dfs(root.right, depth);
        }
    }
}

迭代

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class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int depth = 0;
        while(!queue.isEmpty()){
            int length = queue.size();
            depth++;
            while (length-- > 0){
                TreeNode node = queue.poll();
                if (node.left == null && node.right == null){
                    return depth;
                }

                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }

        return depth;
    }
}

222 完全二叉树节点个数

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class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) return 0;

        TreeNode left = root.left;
        TreeNode right = root.right;
        int leftDepth = 0;
        int rightDepth = 0;

        while (left != null){
            left = left.left;
            leftDepth++;
        }

        while (right != null){
            right = right.right;
            rightDepth++;
        }

        if (leftDepth == rightDepth){
            return (2 << leftDepth) - 1;
        }

        return 1 + countNodes(root.left) + countNodes(root.right);
    }
}

110 平衡二叉树

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class Solution {
    public boolean isBalanced(TreeNode root) {
        return getHeight(root) != -1;
    }

    private int getHeight(TreeNode root){
        if (root == null) return 0;
        int left = getHeight(root.left);
        if (left == -1) return -1;
        int right = getHeight(root.right);
        if (right == -1) return -1;

        if (Math.abs(left - right) > 1){
            return -1;
        }

        return 1 + Math.max(left, right);
    }
}

257 二叉树的所有路径

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class Solution {
    List<String> result = new ArrayList<>();
    public List<String> binaryTreePaths(TreeNode root) {
        if (root == null) return result;
        List<Integer> paths = new ArrayList<>();

        getPaths(root, paths);
        return result;
    }

    private void getPaths(TreeNode root, List<Integer> paths){
        paths.add(root.val);
        if (root.left == null && root.right == null){
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < paths.size() - 1; i++){
                sb.append(paths.get(i)).append("->");
            }
            sb.append(paths.get(paths.size() - 1));

            result.add(sb.toString());

            return;
        }

        if (root.left != null) {
            getPaths(root.left, paths);
            paths.remove(paths.size() - 1);
        }
        if (root.right != null) {
            getPaths(root.right, paths);

            paths.remove(paths.size() - 1);
        }
    }
}

404 左叶子之和

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class Solution {
    int sum = 0;
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
        if (root.left != null && root.left.left == null && root.left.right == null){
            sum += root.left.val;
        }

        if (root.left != null) sumOfLeftLeaves(root.left);
        if (root.right != null) sumOfLeftLeaves(root.right);
        return sum;
    }
}

513 找树左下角的值

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class Solution {
    public int findBottomLeftValue(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int result = 0;
        while(!queue.isEmpty()){
            int length = queue.size();
            for (int i = 0; i < length; i++){
                TreeNode node = queue.poll();
                if (i == 0){
                    result = node.val;
                }

                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        return result;
    }
}
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class Solution {
    int val = 0;
    int maxDepth = 0;
    public int findBottomLeftValue(TreeNode root) {
        if (root == null) return val;
        dfs(root, 0);
        return val;
    }

    private void dfs(TreeNode root, int depth){
        depth++;
        if (root.left == null && root.right == null){
            if (depth > maxDepth){
                maxDepth = depth;
                val = root.val;
            }
        }
        if (root.left != null) dfs(root.left, depth);
        if (root.right != null) dfs(root.right, depth);

    }
}

112 路径总和

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class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) return false;
        targetSum -= root.val;
        if (root.left == null && root.right == null){
            if (targetSum == 0){
                return true;
            }
        }
        boolean left = hasPathSum(root.left, targetSum);
        boolean right = hasPathSum(root.right,targetSum);
        return left || right;
    }
}

113 路径总和II

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {

        if (root == null) return result;
        List<Integer> paths = new ArrayList<>();
        getPaths(root, paths, targetSum);
        return result;
    }

    private void getPaths(TreeNode node, List<Integer> paths, int targetSum){
        paths.add(node.val);
        targetSum -= node.val;
        if (node.left == null && node.right == null){
            if (targetSum == 0){
                result.add(new ArrayList<Integer>(paths));
            }
        }

        if (node.left != null){
            getPaths(node.left, paths, targetSum);
            paths.remove(paths.size() - 1);
        }

        if (node.right != null){
            getPaths(node.right, paths, targetSum);
            paths.remove(paths.size() - 1);
        }
    }
}

106 从中序与后序遍历序列构造二叉树

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class Solution {
    Map<Integer, Integer> map = new HashMap<>();
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder.length == 0) return null;
        for (int i = 0; i < inorder.length; i++){
            map.put(inorder[i], i);
        }
        
        return buildHelper(inorder, 0, inorder.length, postorder, 0, postorder.length);
    }

    private TreeNode buildHelper(int[] inorder, int inS, int inE, int[] postorder, int postS, int postE){
        if (inS >= inE || postS >= postE) return null;

        int rootVal = postorder[postE - 1];
        int rootIndex = map.get(rootVal);
        int lengthOfLeft = rootIndex - inS;
        TreeNode root = new TreeNode(rootVal);

        root.left = buildHelper(inorder, inS, rootIndex, postorder, postS, postS + lengthOfLeft);
        root.right = buildHelper(inorder, rootIndex + 1, inE, postorder, postS + lengthOfLeft, postE - 1);

        return root;
    }
}

105 从前序与中序遍历序列构造二叉树

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class Solution {
    Map<Integer, Integer> map = new HashMap<>();
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder.length == 0) return null;

        for (int i = 0; i < inorder.length; i++){
            map.put(inorder[i], i);
        }

        return buildHelper(preorder, 0, preorder.length, inorder, 0, inorder.length);
    }

    private TreeNode buildHelper(int[] preorder, int preS, int preE, int[] inorder, int inS, int inE){
        if (preS >= preE || inS >= inE) return null;

        int rootVal = preorder[preS];
        int rootIndex = map.get(rootVal);
        int lenOfLeft = rootIndex - inS;
        TreeNode root = new TreeNode(rootVal);

        root.left = buildHelper(preorder, preS + 1, preS + lenOfLeft + 1, inorder, inS, rootIndex);
        root.right = buildHelper(preorder, preS + lenOfLeft + 1, preE, inorder, rootIndex + 1, inE);
        return root;
    }
}

654 最大二叉树

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class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        if (nums.length == 0) return null;
        return constructHelper(nums, 0, nums.length);
    }

    private TreeNode constructHelper(int[] nums, int start, int end){
        if (start >= end) return null;
        
        int[] max = getMaximum(nums, start, end);
        int maxValue = max[0];
        int maxIndex = max[1];

        TreeNode root = new TreeNode(maxValue);
        root.left = constructHelper(nums, start, maxIndex);
        root.right = constructHelper(nums, maxIndex + 1, end);

        return root;

    }

    private int[] getMaximum(int[] nums, int start, int end){
        int maxVal = Integer.MIN_VALUE;
        int index = start;
        for (int i = start; i < end; i++){
            if (nums[i] > maxVal){
                maxVal = nums[i];
                index = i;
            }
        }

        return new int[] {maxVal, index};
    }
}

617 合并二叉树

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class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) return null;

        if (root1 == null && root2 != null) return root2;

        if (root1 != null && root2 == null) return root1;

        TreeNode root = new TreeNode(root1.val + root2.val);
        root.left = mergeTrees(root1.left, root2.left);
        root.right = mergeTrees(root1.right, root2.right);
        return root;
    }
}

700 二叉搜索树中的搜索

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class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if (root == null) return null;
        if (root.val == val){
            return root;
        } else if (root.val > val){
            return searchBST(root.left, val);
        } else {
            return searchBST(root.right, val);
        }
    }
}

98 验证二叉搜索树

中序递归判断数组

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class Solution {
    List<Integer> result = new ArrayList<>();

    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        inorder(root);
        return isInOrder(result);
    }

    private void inorder(TreeNode root){
        if (root == null) return;
        inorder(root.left);
        result.add(root.val);
        inorder(root.right);
    }
    private boolean isInOrder(List<Integer> result){
        for (int i = 1; i < result.size(); i++){
            if (result.get(i - 1) >= result.get(i)) return false;
        }
        return true;
    }
}

记录上一个node 比较root和node的大小

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class Solution {
    TreeNode max;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        boolean left = isValidBST(root.left);
        if (!left) return false;

        if (max != null && root.val <= max.val) return false;
        max = root;

        boolean right = isValidBST(root.right);
        return right;
    }
}

107 二叉树的层次遍历II

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();

        queue.offer(root);

        while(!queue.isEmpty()){
            int length = queue.size();
            List<Integer> item = new ArrayList<>();
            while(length-- > 0){
                TreeNode node = queue.poll();
                item.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            result.add(item);
        }

        Collections.reverse(result);

        return result;
    }
}

199 二叉树的右视图

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class Solution {
    List<Integer> result = new ArrayList<>();
    public List<Integer> rightSideView(TreeNode root) {
        if (root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();

        queue.offer(root);

        while(!queue.isEmpty()){
            int length = queue.size();
            while(length-- > 0){
                TreeNode node = queue.poll();
                if (length == 0){
                    result.add(node.val);
                }

                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        
        return result;
    }
}

530 二叉搜索树的最小绝对差

利用中序遍历二叉搜索树，比较当前值和之前的值的差值

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class Solution {
    List<Integer> result = new ArrayList<>();
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return 0;
        inorder(root);
        int min = Integer.MAX_VALUE;
        for (int i = 1; i < result.size(); i++){
            int diff = result.get(i) - result.get(i -1);
            if (diff < min) {
                min = diff;
            }
        }

        return min;
    }

    private void inorder(TreeNode node){
        if (node == null) return;

        inorder(node.left);
        result.add(node.val);
        inorder(node.right);
    }
}

在中序遍历之中记录一个pre node 之后再记录min的差值，进行比较，二叉搜索树特性

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class Solution {
    int min = Integer.MAX_VALUE;
    TreeNode pre;
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return 0;
        getMinimumDifference(root.left);
        
        if (pre != null && Math.abs(pre.val - root.val) < min) min = Math.abs(pre.val - root.val);
        pre = root;
        getMinimumDifference(root.right);

        return min;
    }
}

501 二叉搜索树中的众数

利用map 前序遍历计数，然后再统计出现频率最多的数字
需要背语法

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class Solution {
    Map<Integer, Integer> map = new HashMap<>();
    List<Integer> result = new ArrayList<>();
    public int[] findMode(TreeNode root) {
        preorder(root);
        List<Map.Entry<Integer, Integer>> mapList = map.entrySet().stream()
				.sorted((c1, c2) -> c2.getValue().compareTo(c1.getValue()))
				.collect(Collectors.toList());
        result.add(mapList.get(0).getKey());
        for (int i = 1; i < mapList.size(); i++) {
			if (mapList.get(i).getValue() == mapList.get(i - 1).getValue()) {
				result.add(mapList.get(i).getKey());
			} else {
				break;
			}
		}
		return result.stream().mapToInt(Integer::intValue).toArray();
    }

    private void preorder(TreeNode root){
        if (root == null) return;
        if (map.containsKey(root.val)){
            map.put(root.val, map.get(root.val) + 1);
        } else {
            map.put(root.val, 1);
        }

        preorder(root.left);
        preorder(root.right);
    }
}

一次中序遍历利用二叉搜索树特性
记录prenode后，记录count， 比较maxcount，等于添加到数组中，大于就清空原来的数组，再添加root

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class Solution {
    List<Integer> result = new ArrayList<>();
    int maxCount = 0;
    int count = 0;
    TreeNode pre;
    public int[] findMode(TreeNode root) {
        inorder(root);
        int[] res = new int[result.size()];
        for (int i = 0; i < result.size(); i++) {
            res[i] = result.get(i);
        }
        return res;
    }

    private void inorder(TreeNode root){
        if (root == null) return;

        inorder(root.left);
        if (pre == null){
            count = 1;
        } else if (pre.val == root.val){
            count++;
        } else {
            count = 1;
        }

        pre = root;

        if (count > maxCount){
            result.clear();
            result.add(root.val);
            maxCount = count;
        } else if (count == maxCount){
            result.add(root.val);
        }

        inorder(root.right);
    }
}

迭代法同理，在处理node时同样的逻辑

236 二叉树的最近公共祖先

此题是自底向上查找，就是回溯
在二叉树中 后序遍历就是自底向上

如果有一个节点，发现左子树出现节点P，右子树出现节点q，或者相反，那么该节点就是公共祖先

如果遇到p 返回p如果遇到q返回q，如果左右子树不为空，中节点就是最近祖先
整体过程如图

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class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q){
            return root;
        }

        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p , q);

        if (left != null && right != null) return root;

        if (left == null && right != null) return right;
        else if (left != null && right == null) return left;
        else return null;
        
    }
}