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

Code

Time Complexity: $O(n)$, Space Complexity: $O(n)$

使用 Binary Tree Level Order Traversal 就行了。

/**
 * 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) {
        queue<TreeNode*> q;
        q.push(root);
        int level = 0;
        while(!q.empty()) {
            int n = q.size();
            int prev = level % 2 == 0 ? 0 : 1e6 + 1;
            for(int i = 0; i < n; i++) {
                bool increasing = level % 2 == 0 ? true : false;
                auto node = q.front();
                q.pop();
 
                if(increasing) {
                    if(node->val % 2 == 0)
                        return false;
                    else {
                        if(node->val <= prev)
                            return false;
                    }
                } else {
                    if(node->val % 2 != 0)
                        return false;
                    else {
                        if(node->val >= prev)
                            return false;
                    }
                }
 
                prev = node->val;
                
                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }
            level++;
        }
        return true;
    }
};

Source

• Even Odd Tree - LeetCode