# [LeetCode] Balanced Binary Tree 平衡二叉树

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 everynode never differ by more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
/ \
9  20
/  \
15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
/ \
2   2
/ \
3   3
/ \
4   4


Return false.

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

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

Maximum Depth of Binary Tree

https://leetcode.com/problems/balanced-binary-tree/

https://leetcode.com/problems/balanced-binary-tree/discuss/35691/The-bottom-up-O(N)-solution-would-be-better

https://leetcode.com/problems/balanced-binary-tree/discuss/35686/Java-solution-based-on-height-check-left-and-right-node-in-every-recursion-to-avoid-further-useless-search

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2014-10-23 13:59  Grandyang  阅读(17418)  评论(7编辑  收藏  举报