leetcode:110 Balanced Binary Tree－每日編程第十九題

Balanced Binary Tree

Total Accepted: 85721 Total Submissions: 261445 Difficulty: Easy

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.

思路：

1).判斷root是否平衡（即左右子樹的高是否相差在1 內）,平衡轉2).，如果不平衡,則返回false;

2).判斷root的左子樹是否平衡，root的右子樹是否平衡，即把root的左右節點當成root並重復1).

3).直到NULL結束。

/**
 * 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 getHeight(TreeNode* root){
        if(root==NULL){
            return 0;
        }else{
            return max(getHeight(root->left),getHeight(root->right))+1;
        }
    }

    bool isBalanced(TreeNode* root) {
        if(root==NULL){
            return true;
        }
        int m = getHeight(root->left);
        int n = getHeight(root->right);
        if(abs(n-m)<=1){
            return isBalanced(root->left)&&isBalanced(root->right);
        }else{
            return false;
        }
    }
};