LeetCode:Balanced Binary Tree(判断是否为二叉平衡树)-优快云博客

本文介绍了一种算法，用于判断二叉树是否为高度平衡的二叉树。通过两次递归的方式，先计算每个节点的左右子树的深度，再判断深度差是否不超过1来确定整棵树是否平衡。

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.

//两次递归效率不高合并看leetcode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=getDepth(root->left);
        int rightDepth=getDepth(root->right);
        if(abs(leftDepth-rightDepth)>1) 
            return false;
        else
            return isBalanced(root->left)&&isBalanced(root->right);
        
    }
    //得到树的高度
    int getDepth(TreeNode *node)
    {
        if(node==NULL)
            return 0;
        int leftDepth=getDepth(node->left);
        int rightDepth=getDepth(node->right);
        return 1+max(leftDepth,rightDepth);
    }
};

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool isBalanced(TreeNode* root) {
13         return balancedHeight(root)>=0;
14         
15     }
16     
17     //return the height of root if the tree is a balanced tree;
18     //otherwise,return -1
19     int balancedHeight(TreeNode *root)
20     {
21         if(root==NULL) return 0;
22         int left=balancedHeight(root->left);
23         int right=balancedHeight(root->right);
24         if(left<0||right<0||abs(left-right)>1)
25             return -1;
26         else
27             return max(left,right)+1;
28     }
29 };