#49 Easy Time O(n) Space O(n) Animated

Balanced Binary Tree

Check if a binary tree is height-balanced.

LeetCode ↗

Tutor Mode Build the mental model before you memorize the code.

1 Predict
2 Trace
3 Explain
4 Implement

Pattern Tree traversal with recursive contracts

Ask each subtree for exactly the information its parent needs, then combine the two answers.

Approach: DFS (Bottom-Up)

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Core Insight

Ask each subtree for exactly the information its parent needs, then combine the two answers.

Common Pitfall

Define the base case before combining left and right subtree results.

Description

Check if a binary tree is height-balanced.

Constraints

Time Complexity: O(n)
Space Complexity: O(n)

Solution — DFS (Bottom-Up)

TimeO(n) SpaceO(n)

Implementation Plan

Translate the invariant into these small, testable moves.

Define what one recursive call returns.
Handle an empty node.
Recurse into children.
Combine the child results at the current node.

def isBalanced(root):
    def dfs(node):
        if not node:
            return [True, 0]
            
        left = dfs(node.left)
        right = dfs(node.right)
        
        balanced = (left[0] and right[0] and 
                    abs(left[1] - right[1]) <= 1)
                    
        return [balanced, 1 + max(left[1], right[1])]
        
    return dfs(root)[0]

Diameter of Binary Tree

Same Tree

Balanced Binary Tree

Steps

Description

Constraints

Tags

Solution — DFS (Bottom-Up)