1609. Even Odd Tree

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 the 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.

Example 4:

Input: root = [1]
Output: true

Example 5:

Input: root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
Output: true

Constraints:

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

class Solution {
    public boolean isEvenOddTree(TreeNode root) {
        if(root == null) return true;
        Queue<TreeNode> q = new LinkedList();
        q.add(root);
        boolean even = true;
        while(q.size() > 0) {
            int size = q.size();
            int prevVal = even ? Integer.MIN_VALUE : Integer.MAX_VALUE; // init preVal based on level even or odd
            while(size-- > 0) { // level by level
                root = q.poll();
                if(even && (root.val % 2 == 0 || root.val <= prevVal)) return false; // invalid case on even level
                if(!even && (root.val % 2 == 1 || root.val >= prevVal)) return false; // invalid case on odd level
                prevVal = root.val; // update the prev value
                if(root.left != null) q.add(root.left); // add left child if exist
                if(root.right != null) q.add(root.right); // add right child if exist
            }
            even = !even; // alter the levels
        }
        
        return true;
    }
}

https://leetcode.com/problems/even-odd-tree/discuss/877723/Java-BFS-Traversal-with-Heavily-Commented level order 拓展。