1609. Even Odd Tree

Problem

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

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isEvenOddTree = function(root) {
    var nodesOfLevel = [root];
    var level = 0;
    while (nodesOfLevel.length) {
        var newNodesOfLevel = [];
        for (var i = 0; i < nodesOfLevel.length; i++) {
            if (level % 2) {
                if (nodesOfLevel[i].val % 2) return false;
                if (i > 0 && nodesOfLevel[i].val >= nodesOfLevel[i - 1].val) return false;
            } else {
                if (nodesOfLevel[i].val % 2 === 0) return false;
                if (i > 0 && nodesOfLevel[i].val <= nodesOfLevel[i - 1].val) return false;
            }
            nodesOfLevel[i].left && newNodesOfLevel.push(nodesOfLevel[i].left);
            nodesOfLevel[i].right && newNodesOfLevel.push(nodesOfLevel[i].right);
        }
        nodesOfLevel = newNodesOfLevel;
        level += 1;
    }
    return true;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).