110. Balanced Binary Tree
https://leetcode.com/problems/balanced-binary-tree/
---
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 7Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
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| # Definition for a binary tree node. | |
| # class TreeNode: | |
| # def __init__(self, val=0, left=None, right=None): | |
| # self.val = val | |
| # self.left = left | |
| # self.right = right | |
| class Solution: | |
| def isBalanced(self, root: Optional[TreeNode]) -> bool: | |
| if not root: | |
| return True | |
| return self.dfs(root)[1] | |
| def dfs(self, node: TreeNode) -> Tuple[int, bool]: | |
| if not node: | |
| return (0, True) | |
| left = self.dfs(node.left) | |
| if not left[1]: | |
| return (0, False) | |
| right = self.dfs(node.right) | |
| if not right[1]: | |
| return (0, False) | |
| if abs(left[0] - right[0]) > 1: | |
| return (0, False) | |
| return (1 + max(left[0], right[0]), True) | |
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| /** | |
| * Definition for a binary tree node. | |
| * public class TreeNode { | |
| * int val; | |
| * TreeNode left; | |
| * TreeNode right; | |
| * TreeNode() {} | |
| * TreeNode(int val) { this.val = val; } | |
| * TreeNode(int val, TreeNode left, TreeNode right) { | |
| * this.val = val; | |
| * this.left = left; | |
| * this.right = right; | |
| * } | |
| * } | |
| */ | |
| class Pair { | |
| int height; | |
| boolean balanced; | |
| public Pair(int h, boolean b) { | |
| height = h; | |
| balanced = b; | |
| } | |
| } | |
| class Solution { | |
| public boolean isBalanced(TreeNode root) { | |
| return dfs(root).balanced; | |
| } | |
| private Pair dfs(TreeNode node) { | |
| if (node == null) { | |
| return new Pair(0, true); | |
| } | |
| Pair left = dfs(node.left); | |
| if (!left.balanced) { | |
| return new Pair(0, false); | |
| } | |
| Pair right = dfs(node.right); | |
| if (!right.balanced) { | |
| return new Pair(0, false); | |
| } | |
| if (Math.abs(left.height - right.height) > 1) { | |
| return new Pair(0, false); | |
| } | |
| return new Pair(1 + Math.max(left.height, right.height), true); | |
| } | |
| } |
