110. Balanced Binary Tree*

110. Balanced Binary Tree*

​​https://leetcode.com/problems/balanced-binary-tree/​​

题目描述

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7
Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4
Return false.

C++ 实现 1

(20210311 更新) 之前写的 ​​C++ 实现 2​​​ 时间复杂度较高, 其实可以使用一个全局变量来优化, 在求高度的过程中来判断是否为平衡树. 具体做法是, 再递归求高度的时候, 使用 ​​res​​​ 记录当前子树是否为平衡二叉树. 思路和 563. Binary Tree Tilt*​ 以及 543. Diameter of Binary Tree* 思路一致.

class Solution {
private:
    bool res = true;
    int height(TreeNode *root) {
        if (!root) return 0;
        auto l = height(root->left);
        auto r = height(root->right);
        res = res && (abs(r - l) <= 1);
        return max(l, r) + 1;
    }
public:
    bool isBalanced(TreeNode* root) {
        height(root);
        return res;
    }
};

C++ 实现 2

要保证左右子树都是 balanced 并且左右子树的高度差不超过 1. 感觉这道题的代码写法和 572. Subtree of Another Tree* 一致.

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