Balanced Binary Tree

Question

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 every node never differ by more than 1.

Example 1:

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

Return true.

Solution -- Recursion

Key point of solution is that we need a helper function to get max height of left child tree and right child tree.

In each level, get height is O(N) time complexity; and considering it's binary tree, level number is logN, so total time complexity is O(NlogN).

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution:
 9     def isBalanced(self, root: TreeNode) -> bool:
10         def getHeight(node: TreeNode) -> int:
11             if not node:
12                 return 0
13             return max(getHeight(node.left), getHeight(node.right)) + 1
14         
15         if not root:
16             return True
17         left_val = getHeight(root.left)
18         right_val = getHeight(root.right)
19         if abs(left_val - right_val) > 1:
20             return False
21         return self.isBalanced(root.left) and self.isBalanced(root.right)

Improved Solution

Evaluate child tree is balanced or not while calculation heights. Time complexity is O(N).

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution:
 9     def isBalanced(self, root: TreeNode) -> bool:
10         def getHeight(node: TreeNode) -> int:
11             if not node:
12                 return 0
13             left = getHeight(node.left)
14             right = getHeight(node.right)
15             if left == -1 or right == -1:
16                 return -1
17             if abs(left - right) > 1:
18                 return -1
19             return max(left, right) + 1
20         
21         if not root:
22             return True
23         return getHeight(root) != -1