LeetCode #1609 — MEDIUM

Even Odd Tree

Move from brute-force thinking to an efficient approach using tree strategy.

Solve on LeetCode

The Problem

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).
For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right).

Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.

Example 1:

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.

Example 2:

Input: root = [5,4,2,3,3,7]
Output: false
Explanation: The node values on each level are:
Level 0: [5]
Level 1: [4,2]
Level 2: [3,3,7]
Node values in level 2 must be in strictly increasing order, so the tree is not Even-Odd.

Example 3:

Input: root = [5,9,1,3,5,7]
Output: false
Explanation: Node values in the level 1 should be even integers.

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
1 <= Node.val <= 10⁶

Step 01

Brute Force Baseline

Problem summary: 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). For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right). Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[1,10,4,3,null,7,9,12,8,6,null,null,2]

Example 2

[5,4,2,3,3,7]

Example 3

[5,9,1,3,5,7]

Step 02

Core Insight

What unlocks the optimal approach

Use the breadth-first search to go through all nodes layer by layer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1609: Even Odd Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isEvenOddTree(TreeNode root) {
        boolean even = true;
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int prev = even ? 0 : 1000001;
            for (int n = q.size(); n > 0; --n) {
                root = q.pollFirst();
                if (even && (root.val % 2 == 0 || prev >= root.val)) {
                    return false;
                }
                if (!even && (root.val % 2 == 1 || prev <= root.val)) {
                    return false;
                }
                prev = root.val;
                if (root.left != null) {
                    q.offer(root.left);
                }
                if (root.right != null) {
                    q.offer(root.right);
                }
            }
            even = !even;
        }
        return true;
    }
}

// Accepted solution for LeetCode #1609: Even Odd Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isEvenOddTree(root *TreeNode) bool {
	even := true
	q := []*TreeNode{root}
	for len(q) > 0 {
		var prev int = 1e7
		if even {
			prev = 0
		}
		for n := len(q); n > 0; n-- {
			root = q[0]
			q = q[1:]
			if even && (root.Val%2 == 0 || prev >= root.Val) {
				return false
			}
			if !even && (root.Val%2 == 1 || prev <= root.Val) {
				return false
			}
			prev = root.Val
			if root.Left != nil {
				q = append(q, root.Left)
			}
			if root.Right != nil {
				q = append(q, root.Right)
			}
		}
		even = !even
	}
	return true
}

# Accepted solution for LeetCode #1609: Even Odd Tree
# 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

// Accepted solution for LeetCode #1609: Even Odd Tree
struct Solution;
use rustgym_util::*;

use std::collections::VecDeque;

impl Solution {
    fn is_even_odd_tree(root: TreeLink) -> bool {
        let mut queue: VecDeque<TreeLink> = VecDeque::new();
        let mut level = 0;
        queue.push_back(root);
        while !queue.is_empty() {
            let n = queue.len();
            let mut nodes = vec![];
            for _ in 0..n {
                if let Some(node) = queue.pop_front().unwrap() {
                    let mut node = node.borrow_mut();
                    nodes.push(node.val);
                    queue.push_back(node.left.take());
                    queue.push_back(node.right.take());
                }
            }
            if nodes.iter().any(|&x| x % 2 == level % 2) {
                return false;
            }
            if nodes
                .windows(2)
                .any(|w| (level % 2 == 0 && w[0] >= w[1]) || (level % 2 == 1 && w[0] <= w[1]))
            {
                return false;
            }
            level += 1;
        }
        true
    }
}

#[test]
fn test() {
    let root = tree!(
        1,
        tree!(10, tree!(3, tree!(12), tree!(8)), None),
        tree!(4, tree!(7, tree!(6), None), tree!(9, None, tree!(2)))
    );
    let res = true;
    assert_eq!(Solution::is_even_odd_tree(root), res);
    let root = tree!(5, tree!(9, tree!(3), tree!(5)), tree!(1, tree!(7), None));
    let res = false;
    assert_eq!(Solution::is_even_odd_tree(root), res);
    let root = tree!(5, tree!(4, tree!(3), tree!(3)), tree!(2, tree!(7), None));
    let res = false;
    assert_eq!(Solution::is_even_odd_tree(root), res);
}

// Accepted solution for LeetCode #1609: Even Odd Tree
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1609: Even Odd Tree
// /**
//  * Definition for a binary tree node.
//  * public class TreeNode {
//  *     int val;
//  *     TreeNode left;
//  *     TreeNode right;
//  *     TreeNode() {}
//  *     TreeNode(int val) { this.val = val; }
//  *     TreeNode(int val, TreeNode left, TreeNode right) {
//  *         this.val = val;
//  *         this.left = left;
//  *         this.right = right;
//  *     }
//  * }
//  */
// class Solution {
//     public boolean isEvenOddTree(TreeNode root) {
//         boolean even = true;
//         Deque<TreeNode> q = new ArrayDeque<>();
//         q.offer(root);
//         while (!q.isEmpty()) {
//             int prev = even ? 0 : 1000001;
//             for (int n = q.size(); n > 0; --n) {
//                 root = q.pollFirst();
//                 if (even && (root.val % 2 == 0 || prev >= root.val)) {
//                     return false;
//                 }
//                 if (!even && (root.val % 2 == 1 || prev <= root.val)) {
//                     return false;
//                 }
//                 prev = root.val;
//                 if (root.left != null) {
//                     q.offer(root.left);
//                 }
//                 if (root.right != null) {
//                     q.offer(root.right);
//                 }
//             }
//             even = !even;
//         }
//         return true;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.