# 1609. Even Odd Tree

## Description

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

## Solutions

BFS.

• /**
* 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) {
Queue < TreeNode > que = new LinkedList < > ();
int res = 1;
int c = 0;
int prev = 0;
while (!que.isEmpty()) {
if (res % 2 == 0)
prev = Integer.MAX_VALUE;
else
prev = Integer.MIN_VALUE;
while (que.peek() != null) {
if (que.peek().val % 2 != res % 2) {
return false;
}
if (res % 2 == 0 && prev <= que.peek().val) {
return false;
}

if (res % 2 != 0 && prev >= que.peek().val)
return false;
if (que.peek().left != null)
if (que.peek().right != null)
prev = que.poll().val;
}
que.poll();
if (que.isEmpty()) {
break;
}
res++;

}
return true;
}
}

• /**
* Definition for a binary tree node.
* 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) {
int even = 1;
queue<TreeNode*> q{ {root} };
while (!q.empty()) {
int prev = even ? 0 : 1e6;
for (int n = q.size(); n; --n) {
root = q.front();
q.pop();
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) q.push(root->left);
if (root->right) q.push(root->right);
}
even ^= 1;
}
return true;
}
};

• # 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


• /**
* 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 = 1e6
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
}