Leetcode 110. Balanced Binary Tree

110. Balanced Binary Tree

• Total Accepted: 119517
• Total Submissions:345415
• 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：
方法一：直接按照定义做。平衡树的定义：左右子树高差的绝对值不大于1且左右子树都是平衡树。

方法二：设计了isDepth函数：如果root为根的树是平衡树，则返回树高；否则返回-1。

代码：
方法一：

class Solution {
public:
    int Depth(TreeNode* root){
        if(!root){
            return 0;
        }
        return 1+max(Depth(root->left),Depth(root->right));
    }
    bool isBalanced(TreeNode* root) {
        if(!root){
            return true;
        }
        int left=Depth(root->left);
        int right=Depth(root->right);
        return abs(left-right)<=1&&isBalanced(root->left)&&isBalanced(root->right);    
    }
};

方法二：

/**
 * 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:
    //如果root为根的树是平衡树，则返回树高；否则返回-1
    int isDepth(TreeNode* root){
        if(!root){
            return 0;
        }
        int left=isDepth(root->left);
        if(left==-1) return -1;
        int right=isDepth(root->right);
        if(right==-1) return -1;
        if(abs(left-right)<=1){
            return max(left,right)+1;
        }
        return -1;
    }
    bool isBalanced(TreeNode* root) {
        return isDepth(root)!=-1;
    }
};