[Leetcode] 110. Balanced Binary Tree 解题报告_110. balanced binary tree magicbean-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/70770128

本文介绍了一种判断二叉树是否平衡的方法。通过递归计算每个节点左右子树的高度差来确定整棵树是否平衡，并提供了具体的C++实现代码。

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题目：

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 depth of the two subtrees of every node never differ by more than 1.

思路：

判断一棵树是否为平衡树可以转换为求解其两棵子树的高度差。一旦高度差大于1，则说明不平衡。但是在递归求解的时候，一旦发现其某棵子树不平衡，则需要提前返回。

代码：

/**
 * 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 {
public:
    bool isBalanced(TreeNode* root) {
        if (root == NULL) {
            return true;
        }
        if (!isBalanced(root->left) || !isBalanced(root->right)) {
            return false;
        }
        int left_height = getHeight(root->left);
        int right_height = getHeight(root->right);
        return abs(left_height - right_height) <= 1;
    }
private:
    int getHeight(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }
        int left_height = getHeight(root->left);
        int right_height = getHeight(root->right);
        return max(left_height, right_height) + 1;
    }
};