How do you solve Even Odd Tree on LeetCode?

Even Odd Tree (LeetCode #1609) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Even-indexed levels must have odd values in increasing order.

What is the brute force solution for Even Odd Tree?

The brute force approach for Even Odd Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. We can traverse the tree level by level and check the values at each level against the Even-Odd conditions. This involves storing all values of each level and then validating them after the traversal. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Even Odd Tree?

Even Odd Tree uses the following concepts: Tree, Breadth-First Search, Binary Tree. The recommended patterns to study are: Breadth-First Search, Tree Traversal.

What are the interview tips for Even Odd Tree?

Clarify the problem requirements before coding. Think about edge cases, like single-node trees. Explain your thought process while coding.

What are common mistakes when solving Even Odd Tree?

Not checking for strict inequalities. Confusing even and odd levels.

#1609

Even Odd Tree

Medium

TreeBreadth-First SearchBinary TreeBreadth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can still use a level-order traversal but check the conditions as we collect values, which avoids the need to store all values before validating them. This reduces unnecessary checks and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Use a queue for level-order traversal of the tree.
2Step 2: For each level, keep track of the previous value to ensure the current values meet the Even-Odd conditions.
3Step 3: Validate the values immediately as we traverse, returning false as soon as a condition fails.

solution.py25 lines

1from collections import deque
2
3def isEvenOddTree(root):
4    if not root:
5        return True
6    queue = deque([root])
7    level = 0
8    while queue:
9        level_size = len(queue)
10        prev_value = None
11        for _ in range(level_size):
12            node = queue.popleft()
13            if level % 2 == 0:
14                if node.val % 2 == 0 or (prev_value is not None and node.val <= prev_value):
15                    return False
16            else:
17                if node.val % 2 != 0 or (prev_value is not None and node.val >= prev_value):
18                    return False
19            prev_value = node.val
20            if node.left:
21                queue.append(node.left)
22            if node.right:
23                queue.append(node.right)
24        level += 1
25    return True

ℹ

Complexity note: The time complexity is O(n) because we traverse each node exactly once. The space complexity is O(n) due to the queue storing nodes at each level.

1Even-indexed levels must have odd values in increasing order.
2Odd-indexed levels must have even values in decreasing order.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.