【leetcode】110. Balanced Binary Tree

一、题目描述

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.

题目解读：判断一棵树是否是平衡二叉树。平衡二叉树满足的条件（1）两边的子树的高度相差不超过1。（2）两棵子树都是平衡二叉树

思路：针对条件（1）要比较高度的话，需要写一个函数求一棵树的高度。

   针对条件（2）很容易就想到可以用递归的方法

c++代码（16ms，14.24%）

/**
* 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 height(TreeNode* root){
//求一棵树的高度
if(root == NULL)
return 0;
else if((root->left == NULL) && (root->right == NULL))
return 1;
else
return max(height(root->left), height(root->right))+1;
}

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

思路一样，代码可以进一步简化为：

class solution {
public:
int depth (TreeNode *root) {
if (root == NULL) return 0;
return max (depth(root -> left), depth (root -> right)) + 1;
}

bool isBalanced (TreeNode *root) {
if (root == NULL) return true;

int left=depth(root->left);
int right=depth(root->right);

return abs(left - right) <= 1 && isBalanced(root->left) && isBalanced(root->right);
}
};

代码2：

class solution {
public:
int dfsHeight (TreeNode *root) {
if (root == NULL) return 0;

int leftHeight = dfsHeight (root -> left);
if (leftHeight == -1) return -1;
int rightHeight = dfsHeight (root -> right);
if (rightHeight == -1) return -1;

if (abs(leftHeight - rightHeight) > 1) return -1;
return max (leftHeight, rightHeight) + 1;
}
bool isBalanced(TreeNode *root) {
return dfsHeight (root) != -1;
}
};