LeetCode 110 Balanced Binary Tree

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于 2018-08-13 11:38:46 发布

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本文链接：https://blog.youkuaiyun.com/yige__cxy/article/details/81627077

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本文介绍了一种用于判断二叉树是否为高度平衡的算法。通过递归计算每个节点的左右子树深度，并确保任意节点的左右子树深度差不超过1来实现平衡性的检查。文章提供了完整的C++代码实现。

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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.

Example 1:

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

Return true.

Example 2:

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

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;
        int leftdepth = depth(root->left);
        int rightdepth = depth(root->right);
        if(abs(leftdepth-rightdepth)>=2)//这一步优化很重要，运行时间可以从8ms减少到4ms
            return false;
        else
            return isBalanced(root->left) && isBalanced(root->right);
    }
    int depth(TreeNode* root){
        if(root == NULL) return 0;
        return max(depth(root->left),depth(root->right))+1;
    }

};