LeetCode 110. Balanced Binary Tree
原题链接在这里：https://leetcode.com/problems/balanced-binary-tree/
题目：
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 left and right subtrees of every node differ in height by no 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.
题解：
Get 最大深度.如果左右最大深度相差大于一，则返回-1. 若是已有左或者右的返回值为-1, 即返回-1.
最后看返回到root时是否为一个负数，若是负数则不是balanced, 若是正数，则返回了最大深度，是balanced.
Time Complexity: O(n).
Space: O(log n), n is number of nodes in the tree.
AC Java:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public boolean isBalanced(TreeNode root) { 12 return maxDepth(root) >= 0; 13 } 14 15 private int maxDepth(TreeNode root){ 16 if(root == null){ 17 return 0; 18 } 19 int leftDepth = maxDepth(root.left); 20 int rightDepth = maxDepth(root.right); 21 if(leftDepth == -1 || rightDepth == -1 || Math.abs(leftDepth - rightDepth) > 1){ 22 return -1; 23 } 24 return Math.max(leftDepth, rightDepth) + 1; 25 } 26 }
AC C++:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 bool isBalanced(TreeNode* root) { 15 return maxDepth(root) >= 0; 16 } 17 18 int maxDepth(TreeNode* root){ 19 if(!root){ 20 return 0; 21 } 22 23 int l = maxDepth(root->left); 24 int r = maxDepth(root->right); 25 if(l < 0 || r < 0 || abs(l - r) > 1){ 26 return -1; 27 } 28 29 return max(l, r) + 1; 30 } 31 };
