代码随想录第十九天

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复习

深度优先遍历

前中后序遍历

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);
    }
}

统一迭代

后序为例

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

        Stack<TreeNode> stack = new Stack<>();

        stack.push(root);

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

        return result;
    }
}

不同迭代

前序

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class Solution {
    List<Integer> result = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        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 {
    List<Integer> result = new ArrayList<>();
    public List<Integer> inorderTraversal(TreeNode root) {
        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 {
    List<Integer> result = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        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;
    }
}

DFS

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class Solution {
    List<List<Integer>> result = new ArrayList<>();
    public List<List<Integer>> levelOrder(TreeNode root) {
        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 {
    List<List<Integer>> result = new ArrayList<>();
    public List<List<Integer>> levelOrder(TreeNode root) {
        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;
    } 
}

翻转二叉树

前序遍历

递归

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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;
    }

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

迭代

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class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return root;
        Stack<TreeNode> stack = new Stack<>();

        stack.push(root);

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

        return root;
    }

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

对称二叉树、

前序
递归

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class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return false;
        return compare(root.left, root.right);
    }

    public 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;
    }
}

迭代

利用队列

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

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);
        while (!queue.isEmpty()){
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();

            if (left == null && right == null) continue;
            if (left == null || right == null || left.val != right.val) return false;

            queue.offer(left.left);
            queue.offer(right.right);
            queue.offer(left.right);
            queue.offer(right.left);
        }

        return true;
    }
}

二叉树最大深度

递归
后序 高度
前序 深度

此题求根节点高度

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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));
    }
}

迭代层序

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

        Queue<TreeNode> queue = new LinkedList<>();
        int depth = 0;
        queue.offer(root);

        while (!queue.isEmpty()){
            int length = queue.size();
            while (length-- > 0){
                TreeNode node = queue.poll();
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            depth++;
        }

        return depth;
    }
}

二叉树最小深度

迭代 层序遍历，遇到叶子结点直接返回

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class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> queue = new LinkedList<>();
        int depth = 0;
        queue.offer(root);
        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;
    }
}

DFS
慢是因为会遍历所有结点，但层序遍历不会

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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;
    }

    public void dfs(TreeNode root, int depth){
        if (root == null) return;

        depth++;
        if (root.left == null && root.right == null){
            minDepth = Math.min(depth, minDepth);
            return;
        }
        if (root.left != null) dfs(root.left, depth);
        if (root.right != null) dfs(root.right, depth);
    }
}

完全二叉树的结点个数

递归

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

完全二叉树性质

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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);
    }
}

平衡二叉树

高度 后序

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

    public int getHeight(TreeNode node){
        if (node == null) return 0;

        int leftHeight = getHeight(node.left);
        if (leftHeight == -1) return -1;
        int rightHeight = getHeight(node.right);
        if (rightHeight == -1) return -1;

        if (Math.abs(leftHeight - rightHeight) > 1) return -1;
        return 1 + Math.max(leftHeight, rightHeight);
    }
}

二叉树所有路径

前序遍历找路径

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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);
        }
    }
}

左叶子之和

前序遍历，递归

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

        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;
    }
}

找树左下角的值

层序遍历

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

        int result = 0;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        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;
    }
}

dfs 递归记录depth对比

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class Solution {
    int maxDepth = 0;
    int result = 0;

    public int findBottomLeftValue(TreeNode root) {
        if (root == null) return result;
        getValue(root, 0);
        
        return result;    
    }

    public void getValue(TreeNode root, int depth){
        if (root == null) return;
        depth++;

        if (root.left == null && root.right == null){
            if (maxDepth < depth){
                maxDepth = depth;
                result = root.val;
            }
        }

        if (root.left != null) getValue(root.left, depth);
        if (root.right != null) getValue(root.right, depth);
    }
}

路径总和

递归求和

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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 = false;
        boolean right = false;
        if (root.left != null){
            left = hasPathSum(root.left, targetSum);
        }
        if (root.right != null) {
            right = hasPathSum(root.right, targetSum);
        }
        return left || right;
    }
}

路径总和II

递归记录路径

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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, targetSum, paths);

        return result;
    }

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

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

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);
    }
    public 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;
    }
}

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);
    }

    public 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];
        TreeNode root = new TreeNode(rootVal);
        int middleIndex = map.get(rootVal);
        int lenOfLeft = middleIndex - inS;

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