LeetCode:Balanced Binary Tree(判断是否为二叉平衡树)

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.

//两次递归效率不高 合并看leetcode代码

/**
* 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:
bool isBalanced(TreeNode* root) {
if(root==NULL) return true;
int leftDepth=getDepth(root->left);
int rightDepth=getDepth(root->right);
if(abs(leftDepth-rightDepth)>1)
return false;
else
return isBalanced(root->left)&&isBalanced(root->right);

}
//得到树的高度
int getDepth(TreeNode *node)
{
if(node==NULL)
return 0;
int leftDepth=getDepth(node->left);
int rightDepth=getDepth(node->right);
return 1+max(leftDepth,rightDepth);
}
};

/**
* 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:
bool isBalanced(TreeNode* root) {
return balancedHeight(root)>=0;

}

//return the height of root if the tree is a balanced tree;
//otherwise,return -1
int balancedHeight(TreeNode *root)
{
if(root==NULL) return 0;
int left=balancedHeight(root->left);
int right=balancedHeight(root->right);
if(left<0||right<0||abs(left-right)>1)
return -1;
else
return max(left,right)+1;
}
};