Formatted question description: https://leetcode.ca/all/1609.html

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/

// 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;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode() {}
     *     TreeNode(int val) { this.val = val; }
     *     TreeNode(int val, TreeNode left, TreeNode right) {
     *         this.val = val;
     *         this.left = left;
     *         this.right = right;
     *     }
     * }
     */
    class Solution {
        public boolean isEvenOddTree(TreeNode root) {
            if (root == null)
                return false;
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            int level = 0;
            while (!queue.isEmpty()) {
                int size = queue.size();
                int prev = level % 2 == 0 ? Integer.MIN_VALUE : Integer.MAX_VALUE;
                for (int i = 0; i < size; i++) {
                    TreeNode node = queue.poll();
                    int value = node.val;
                    if (level % 2 == value % 2)
                        return false;
                    if (level % 2 == 0 && value <= prev || level % 2 == 1 && value >= prev)
                        return false;
                    prev = value;
                    TreeNode left = node.left, right = node.right;
                    if (left != null)
                        queue.offer(left);
                    if (right != null)
                        queue.offer(right);
                }
                level++;
            }
            return true;
        }
    }
    
  • // OJ: https://leetcode.com/problems/even-odd-tree/
    // 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;
        }
    };
    
  • # 1609. Even Odd Tree
    # https://leetcode.com/problems/even-odd-tree/
    
    # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution:
        def isEvenOddTree(self, root: TreeNode) -> bool:
            if not root: return False
            
            q = deque()
            q.append(root)
            isEven = True
            while q:
                
                n = len(q)
                check = []
                prev = None
                while n:
                    node = q.popleft()
                    val = node.val
                    if isEven:
                        if val % 2 == 0 or prev and val <= prev:
                            return False
                    else:
                        if val % 2 or prev and val >= prev:
                            return False
                    
                    prev = val
                    
                    for t in (node.left,node.right):
                        if t:
                            q.append(t)
                    
                    n -= 1
                    
                isEven = not isEven
                    
            return True
    

All Problems

All Solutions