1609. Even Odd Tree

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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:

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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:

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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:

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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, 10^5].
  • 1 <= Node.val <= 10^6

Hints/Notes

  • binary tree

Solution

Language: C++

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/**
 * 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 m = 1;
        queue<TreeNode*> q;
        if (root)
            q.push(root);
        while (!q.empty()) {
            int n = q.size();
            int val = m ? INT_MIN : INT_MAX;
            for (int i = 0; i < n; i++) {
                TreeNode* cur = q.front();
                q.pop();
                if (cur->val % 2 != m) {
                    return false;
                }
                if (m && cur->val <= val) {
                    return false;
                }
                if (!m && cur->val >= val) {
                    return false;
                }
                val = cur->val;
                if (cur->left)
                    q.push(cur->left);
                if (cur->right)
                    q.push(cur->right);
            }
            m ^= 1;
        }
        return true;
    }
};