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110. Balanced Binary Tree

Difficulty Topics

Description

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:

Input: root = [3,9,20,null,null,15,7]
Output: true

Example 2:

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false

Example 3:

Input: root = []
Output: true

 

Constraints:

  • The number of nodes in the tree is in the range [0, 5000].
  • -104 <= Node.val <= 104

Solution

balanced-binary-tree.py
# 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:
        res = True

        def go(node):
            nonlocal res

            if not node: return 0

            l, r = go(node.left), go(node.right)

            if abs(l - r) > 1:
                res = False

            return 1 + max(l, r)

        go(root)

        return res
balanced-binary-tree.cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int dfs(TreeNode* root){
        if (root == nullptr) return 0;

        int left = dfs(root->left);
        int right = dfs(root->right);

        if (left == -1 || right == -1 || abs(left-right) > 1) return -1;

        return 1 + max(left, right);
    }

    bool isBalanced(TreeNode* root) {
        return dfs(root) != -1;
    }
};