代码随想录第二十天

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复习

深度优先遍历
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;
    }
}

统一迭代法
前序为例

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

        return result;
    }
}

递归
前序为例

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class Solution {
    List<Integer> result = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        preorder(root);
        return result;
    }
    public void preorder(TreeNode node){
        if (node == null) return;
        result.add(node.val);
        preorder(node.left);
        preorder(node.right);
    }
}

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;
    }
    public void dfs(TreeNode root, int depth){
        if (root == null) return;
        depth++;
        if (result.size() < depth){
            result.add(new ArrayList<Integer>());
        }

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

        if (root.left != null) dfs(root.left, depth);
        if (root.right != null) dfs(root.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 null;
        swapChildren(root);
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }

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

101 对称二叉树

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class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return false;

        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 node, int depth){
        if (node == null) return;
        depth++;
        if (node.left == null && node.right == null){
            if (depth < minDepth){
                minDepth = depth;
            }
        }

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

层序迭代

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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) {
        if (root == null) return true;
        return getHeight(root) != -1;
    }

    private int getHeight(TreeNode node){
        if (node == null) return 0;
        int left = getHeight(node.left);
        if (left == -1) return -1;
        int right = getHeight(node.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 node, List<Integer> paths){
        paths.add(node.val);

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

        if (node.right != null) {
            getPaths(node.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 sum;
        getSums(root, 0);

        return sum;
    }

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

        if (node.left != null && node.left.left == null && node.left.right == null){
            sum += node.left.val;
        }
        if (node.left != null) getSums(node.left, depth);
        if (node.right != null) getSums(node.right, depth);
    }
}

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 = root.val;
        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;
    }
}

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));
                return;
            }
        }

        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];
        TreeNode root = new TreeNode(rootVal);
        int rootIndex = map.get(rootVal);
        int lenOfLeft = rootIndex - inS;
        root.left = buildHelper(inorder, inS, rootIndex, postorder, postS, postS + lenOfLeft);
        root.right = buildHelper(inorder, rootIndex + 1, inE, postorder, postS + lenOfLeft, 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 + 1 + lenOfLeft, inorder, inS, rootIndex);
        root.right = buildHelper(preorder, preS + 1 + lenOfLeft, preE, inorder, rootIndex + 1, inE);

        return root;

    }
}


654 最大二叉树

构造树采用前序遍历

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class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return traversal(nums, 0, nums.length);
    }

    private TreeNode traversal(int[] nums, int start, int end){
        if (start >= end) return null;
        int maxValue = Integer.MIN_VALUE;
        int maxIndex = start;
        for (int i = start; i < end; i++){
            if (nums[i] > maxValue){
                maxValue = nums[i];
                maxIndex = i;
            }
        }

        TreeNode root = new TreeNode(maxValue);

        root.left = traversal(nums, start, maxIndex);
        root.right = traversal(nums, maxIndex + 1, end);
        return root;
    }
}

617 合并二叉树

前序遍历

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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

二叉树递归 利用特性

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class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if (root == null) return null;
        if (root.val == val) return root;

        if (root.val < val) return searchBST(root.right, val);
        if (root.val > val) return searchBST(root.left, val);

        return root;
    }
}

二叉树特性的迭代
不需要栈

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

        return root;
    }
}

98 验证二叉搜索树

统一迭代，用一个pre Node记录上一个node进行比较

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class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) return false;

        Stack<TreeNode> stack = new Stack<>();
        TreeNode pre = null;
        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 {
                TreeNode temp = stack.pop();
                if (pre != null && pre.val >= temp.val){
                    return false;
                }

                pre = temp;
            }
        }
        return true;
    }
}

递归(中序遍历) 需要强化记忆

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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;
    }
}