檢測一個樹是否平衡,不需要求出高度,而是從底到頂檢測是否平衡,這樣纔算法時間複雜度爲O(n)。但是需要額外的O(logn)的空間
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root) {
return checkBalance(root) >= 0;
}
int checkBalance(TreeNode *root){
if(root == NULL) return 0;
int left = checkBalance(root->left);
int right = checkBalance(root->right);
if(left < 0 || right < 0 || abs(left-right) > 1) return -1;
return max(left,right) +1;
}
};