FXTi's blog
0%

balanced-binary-tree

Posted on Edited on

Balanced Binary Tree – LeetCode 110

Problem

Description

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.

Answer

Original

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * 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:
    int depth(TreeNode *node){
        if(!node) return 0;
        return 1 + max(depth(node->left),depth(node->right));
    }
    
    bool isBalanced(TreeNode* root) {
        if(!root) return true;
        return abs(depth(root->left) - depth(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right);
    }
};

思路

依旧是DFS递归解，因为BFS在此处需要连续两层状态信息所以不好用，时间复杂度$O(n^2)$，空间复杂度$O(n)$。
耗时$9$ ms，排名$74.26\%$

Better

思路

简化版本，使用-1作为状态码，在计算深度的过程中检测是否平衡，有效避免额外运算。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$9$ ms，排名$74.26\%$

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
/**
 * 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 (checkDepth(root) == -1) return false;
        else return true;
    }
    int checkDepth(TreeNode *root) {
        if (!root) return 0;
        int left = checkDepth(root->left);
        if (left == -1) return -1;
        int right = checkDepth(root->right);
        if (right == -1) return -1;
        int diff = abs(left - right);
        if (diff > 1) return -1;
        else return 1 + max(left, right);
    }
};