Even Odd Tree

🧩 Problem Statement

A binary tree is named Even-Odd if it meets the following conditions:

1. 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.
2. For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).
3. 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

🔍 Approaches

1. BFS with Level Tracking ($O(N)$)

• Concept: Use Breadth-First Search (BFS) to traverse level by level.
• Constraints Check:
• For each level index level:
• Even Level: vals must be odd and increasing.
• Odd Level: vals must be even and decreasing.
• Algorithm:
• BFS Queue. Track level_index.
• For each level, keep track of prev_val.
• Check parity: val % 2 must match expected (odd for even level, even for odd level).
• Check order: Compare val with prev_val (increasing/decreasing).

⏳ Time & Space Complexity

• Time Complexity: $O(N)$.
• Space Complexity: $O(N)$.

🚀 Code Implementations

C++

#include <vector>
#include <queue>
#include <iostream>
#include <climits>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
class Solution {
public:
bool isEvenOddTree(TreeNode* root) {
if (!root) return true;
queue<TreeNode*> q;
q.push(root);
int level = 0;
while (!q.empty()) {
int size = q.size();
int prev_val = (level % 2 == 0) ? INT_MIN : INT_MAX;
for (int i = 0; i < size; i++) {
TreeNode* node = q.front();
q.pop();
// Check parity
if (level % 2 == 0) { // Even level: odd vals, increasing
if (node->val % 2 == 0 || node->val <= prev_val) return false;
} else { // Odd level: even vals, decreasing
if (node->val % 2 != 0 || node->val >= prev_val) return false;
}
prev_val = node->val;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
level++;
}
return true;
}
};

Python

from typing import Optional
from collections import deque
import math
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:
if not root:
return True
queue = deque([root])
level = 0
while queue:
size = len(queue)
prev_val = -math.inf if level % 2 == 0 else math.inf
for _ in range(size):
node = queue.popleft()
if level % 2 == 0: # Even level
if node.val % 2 == 0 or node.val <= prev_val:
return False
else: # Odd level
if node.val % 2 != 0 or node.val >= prev_val:
return False
prev_val = node.val
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
level += 1
return True

Java

import java.util.*;
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) {
if (root == null) return true;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int level = 0;
while (!queue.isEmpty()) {
int size = queue.size();
int prevVal = (level % 2 == 0) ? Integer.MIN_VALUE : Integer.MAX_VALUE;
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (level % 2 == 0) { // Even level: odd vals, increasing
if (node.val % 2 == 0 || node.val <= prevVal) return false;
} else { // Odd level: even vals, decreasing
if (node.val % 2 != 0 || node.val >= prevVal) return false;
}
prevVal = node.val;
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
level++;
}
return true;
}
}

🌍 Real-World Analogy

Strict Dress Code Party:

A party with multiple floors. Floor 0 (Ground) requires people to wear Odd colors and line up by height (Short -> Tall). Floor 1 requires Even colors and line up Tall -> Short. Security checks every floor strictly.

🎯 Summary

✅ Level-wise check: Validating both parity and sorting order at each step.