1609

1609. Even Odd Tree (Medium)

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

Related Topics:
Tree

Solution 1. Level-order traversal

// OJ: https://leetcode.com/problems/even-odd-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool isEvenOddTree(TreeNode* root) {
        queue<TreeNode*> q;
        q.push(root);
        int lv = 0;
        while (q.size()) {
            int cnt = q.size(), prev = lv % 2 == 0 ? 0 : INT_MAX;
            while (cnt--) {
                auto node = q.front();
                q.pop();
                if (lv % 2 == 0) {
                    if (node->val % 2 == 0 || node->val <= prev) return false;
                } else {
                    if (node->val % 2 || node->val >= prev) return false;
                }
                prev = node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            ++lv;
        }
        return true;
    }
};