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balancedBinaryTree.cpp
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balancedBinaryTree.cpp
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// Source : https://oj.leetcode.com/problems/balanced-binary-tree/
// Author : Hao Chen
// Date : 2014-06-28
/**********************************************************************************
*
* 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.
*
*
**********************************************************************************/
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root) {
int height=0;
return isBalancedUtil(root, height);
}
bool isBalancedUtil(TreeNode* root, int& height){
if(root==NULL){
height=0;
return true;
}
int lh=0, rh=0;
bool isLeft = isBalancedUtil(root->left, lh);
bool isRight = isBalancedUtil(root->right, rh);
height = (lh > rh ? lh : rh) + 1;
return (abs(lh-rh)<=1 && isLeft && isRight);
}
};
//Notes:
// I think the above solution should be more efficent than the below,
// but for leetcode, the below solution needs 60ms, the above needs 88ms
class Solution {
public:
bool isBalanced(TreeNode *root) {
if (root==NULL) return true;
int left = treeDepth(root->left);
int right = treeDepth(root->right);
if (left-right>1 || left-right < -1) {
return false;
}
return isBalanced(root->left) && isBalanced(root->right);
}
int treeDepth(TreeNode *root) {
if (root==NULL){
return 0;
}
int left=1, right=1;
left += treeDepth(root->left);
right += treeDepth(root->right);
return left>right?left:right;
}
};