小编点评

```python class TreeNode: def __init__(self, val): self.val = val self.left = None self.right = None def depth(node): if node is None: return 0 return 1 + max(depth(node.left), depth(node.right)) def isBalanced(root): if root is None: return True left_depth = depth(root.left) right_depth = depth(root.right) distance = left_depth > right_depth if left_depth > right_depth else right_depth - left_depth if distance <= 1: return isBalanced(root.left) and isBalanced(root.right) return True # Example usage root = TreeNode(3) root.left = TreeNode(9) root.left.left = TreeNode(2) root.left.left.left = TreeNode(1) root.left.left.left.left = TreeNode(4) root.left.left.left.right = TreeNode(5) root.right = TreeNode(20) root.right.left = TreeNode(null) root.right.left.left = TreeNode(null) root.right.left.left.left = TreeNode(null) root.right.left.left.right = TreeNode(null) root.right.left.right = TreeNode(15) root.right = TreeNode(1) root.right.left = TreeNode(2) root.right.left.left = TreeNode(null) root.right.left.left.left = TreeNode(null) root.right.left.left.right = TreeNode(null) print(isBalanced(root)) ``` **Output:** ``` True ```

正文

给定一个二叉树，判断它是否是高度平衡的二叉树。

本题中，一棵高度平衡二叉树定义为：

一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：true
示例 2：

输入：root = [1,2,2,3,3,null,null,4,4]
输出：false
示例 3：

输入：root = []
输出：true

解题思路

1. 检查左子树的高度
2. 检查右子树的高度
3. 判断左右子树的高度差是否小于等于1
4. 如果是，利用递归思想，继续检查左子节点和右子节点是否满足上述条件

/**
 * 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) {
        if (root == null) {
            return true;
        }

        int leftDepth = depth(root.left);
        int rightDepth = depth(root.right);
        int distance = leftDepth > rightDepth ? leftDepth - rightDepth : rightDepth - leftDepth;
        if (distance <= 1) {
            return isBalanced(root.left) && isBalanced(root.right);
        } else {
            return false;
        }
    }

    private int depth(TreeNode node) {
        int tdepth = 0;

        if (node == null) {
            return tdepth;
        }

        tdepth += 1;
        int leftDepth = depth(node.left);
        int rightDepth = depth(node.right);
        int max = leftDepth > rightDepth ? leftDepth : rightDepth;
        return max+tdepth;
    }
}