代码随想录训练营第三十期|第十七天|二叉树part04|110.平衡二叉树 ● 257. 二叉树的所有路径 ● 404.左叶子之和

110. 平衡二叉树 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
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 Math.max(left, right) + 1;
    }
}

257. 二叉树的所有路径 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List binaryTreePaths(TreeNode root) {
        List res = new ArrayList<>();
        if (root == null) return res;
        List path = new ArrayList<>();
        dfs(root, res, path);
        return res;
    }

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

        if (root.left != null) {
            dfs(root.left, res, path);
            path.remove(path.size() - 1);
        }
        if (root.right != null) {
            dfs(root.right, res, path);
            path.remove(path.size() - 1);
        }
    }
}

404. 左叶子之和 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
        if (root.left == null && root.right == null) return 0;
        int left = sumOfLeftLeaves(root.left);
        if (root.left != null && root.left.left == null && root.left.right == null) {
            left += root.left.val;
        }
        int right = sumOfLeftLeaves(root.right);
        int res = left + right;
        return res;
    }
}