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110. 平衡二叉树

English Version

题目描述

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

 

提示：

  • 树中的节点数在范围 [0, 5000] 内
  • -104 <= Node.val <= 104

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def height(root):
            if not root:
                return 0
            return 1 + max(height(root.left), height(root.right))
        if not root:
            return True
        left_height, right_height = height(root.left), height(root.right)
        return abs(left_height - right_height) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)

Java

/**
 * 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 leftHeight = height(root.left);
        int rightHeight = height(root.right);
        return Math.abs(leftHeight - rightHeight) <= 1 && isBalanced(root.left) && isBalanced(root.right);
    }

    private int height(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = height(root.left);
        int r = height(root.right);
        return 1 + Math.max(l, r);
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
 var isBalanced = function(root) {
    if (root == null) return true; 
    let left = root.left;
    let right = root.right;
    return isBalanced(left) && isBalanced(right) && Math.abs(depth(left) - depth(right)) <= 1;
};

function depth(root) {
    if (root == null) return 0;
    return Math.max(depth(root.left), depth(root.right)) + 1;
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if (root == nullptr) {
            return true;
        }
        int leftHeight = height(root->left);
        int rightHeight = height(root->right);
        return abs(leftHeight - rightHeight) <= 1 && isBalanced(root->left) && isBalanced(root->right);
    }
private:
    int height(TreeNode* root) {
        if (root == nullptr) {
            return 0;
        }
        int l = height(root->left);
        int r = height(root->right);
        return 1 + max(l, r);
    }
};

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