Leetcode #1609: Even Odd Tree
In this guide, we solve Leetcode #1609 Even Odd Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A binary tree is named Even-Odd if it meets the following conditions: The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2, etc. For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Breadth-First Search, Binary Tree
Intuition
We need level-by-level exploration or shortest steps, which is ideal for BFS.
A queue naturally models the frontier of the search.
Approach
Push initial nodes into a queue and expand in layers.
Track visited nodes to prevent cycles.
Steps:
- Initialize queue with start nodes.
- Process level by level.
- Track visited nodes.
Example
Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
Output: true
Explanation: The node values on each level are:
Level 0: [1]
Level 1: [10,4]
Level 2: [3,7,9]
Level 3: [12,8,6,2]
Since levels 0 and 2 are all odd and increasing and levels 1 and 3 are all even and decreasing, the tree is Even-Odd.
Python Solution
# 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 isEvenOddTree(self, root: Optional[TreeNode]) -> bool:
even = 1
q = deque([root])
while q:
prev = 0 if even else inf
for _ in range(len(q)):
root = q.popleft()
if even and (root.val % 2 == 0 or prev >= root.val):
return False
if not even and (root.val % 2 == 1 or prev <= root.val):
return False
prev = root.val
if root.left:
q.append(root.left)
if root.right:
q.append(root.right)
even ^= 1
return True
Complexity
The time complexity is , and the space complexity is , where is the number of nodes in the binary tree. The space complexity is , where is the number of nodes in the binary tree.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.