【easy】110. Balanced Binary Tree判断二叉树是否平衡

判断二叉树是否平衡

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.

以下解法为什么时间复杂度为O（n）？

/**
 * 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 cntHeight(TreeNode *root) {
        if(root == NULL) return ;
        int l = cntHeight(root->left);
        int r = cntHeight(root->right);
        if(l <  || r <  || abs(l-r) > ) return -; //自定义 return -1，表示不平衡的情况
        else  return max(l, r) + ;     //*******
    }
    bool isBalanced(TreeNode *root) {
        if(root == NULL) return true;
        int l = cntHeight(root->left);      //-1表示不平衡
        int r = cntHeight(root->right);     //-1表示不平衡
        if(l <  || r <  || abs(l-r) > ) return false;
        else return true;
    }
};