76 lines
2.0 KiB
C++
76 lines
2.0 KiB
C++
// Source : https://oj.leetcode.com/problems/balanced-binary-tree/
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// Author : Hao Chen
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// Date : 2014-06-28
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/**********************************************************************************
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*
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* Given a binary tree, determine if it is height-balanced.
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*
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* For this problem, a height-balanced binary tree is defined as a binary tree in which
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* the depth of the two subtrees of every node never differ by more than 1.
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*
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*
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**********************************************************************************/
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/**
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* Definition for binary tree
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool isBalanced(TreeNode *root) {
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int height=0;
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return isBalancedUtil(root, height);
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}
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bool isBalancedUtil(TreeNode* root, int& height){
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if(root==NULL){
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height=0;
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return true;
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}
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int lh=0, rh=0;
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bool isLeft = isBalancedUtil(root->left, lh);
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bool isRight = isBalancedUtil(root->right, rh);
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height = (lh > rh ? lh : rh) + 1;
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return (abs(lh-rh)<=1 && isLeft && isRight);
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}
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};
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//Notes:
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// I think the above solution should be more efficent than the below,
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// but for leetcode, the below solution needs 60ms, the above needs 88ms
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class Solution {
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public:
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bool isBalanced(TreeNode *root) {
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if (root==NULL) return true;
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int left = treeDepth(root->left);
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int right = treeDepth(root->right);
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if (left-right>1 || left-right < -1) {
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return false;
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}
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return isBalanced(root->left) && isBalanced(root->right);
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}
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int treeDepth(TreeNode *root) {
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if (root==NULL){
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return 0;
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}
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int left=1, right=1;
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left += treeDepth(root->left);
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right += treeDepth(root->right);
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return left>right?left:right;
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}
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};
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