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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;
    }
};
  • /**
     * 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;
        }
    }
    
    ############
    
    /**
     * 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) {
            boolean even = true;
            Deque<TreeNode> q = new ArrayDeque<>();
            q.offer(root);
            while (!q.isEmpty()) {
                int prev = even ? 0 : 1000000;
                for (int n = q.size(); n > 0; --n) {
                    root = q.pollFirst();
                    if (even && (root.val % 2 == 0 || prev >= root.val)) {
                        return false;
                    }
                    if (!even && (root.val % 2 == 1 || prev <= root.val)) {
                        return false;
                    }
                    prev = root.val;
                    if (root.left != null) {
                        q.offer(root.left);
                    }
                    if (root.right != null) {
                        q.offer(root.right);
                    }
                }
                even = !even;
            }
            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;
        }
    };
    
  • # 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: Optional[TreeNode]) -> bool:
            even = 1
            q = deque([root])
            while q:
                prev = 0 if even else inf
                for _ in range(len(q)):
                    root = q.popleft()
                    if even and (root.val % 2 == 0 or prev >= root.val):
                        return False
                    if not even and (root.val % 2 == 1 or prev <= root.val):
                        return False
                    prev = root.val
                    if root.left:
                        q.append(root.left)
                    if root.right:
                        q.append(root.right)
                even ^= 1
            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
    
  • /**
     * Definition for a binary tree node.
     * type TreeNode struct {
     *     Val int
     *     Left *TreeNode
     *     Right *TreeNode
     * }
     */
    func isEvenOddTree(root *TreeNode) bool {
    	even := true
    	q := []*TreeNode{root}
    	for len(q) > 0 {
    		var prev int = 1e6
    		if even {
    			prev = 0
    		}
    		for n := len(q); n > 0; n-- {
    			root = q[0]
    			q = q[1:]
    			if even && (root.Val%2 == 0 || prev >= root.Val) {
    				return false
    			}
    			if !even && (root.Val%2 == 1 || prev <= root.Val) {
    				return false
    			}
    			prev = root.Val
    			if root.Left != nil {
    				q = append(q, root.Left)
    			}
    			if root.Right != nil {
    				q = append(q, root.Right)
    			}
    		}
    		even = !even
    	}
    	return true
    }
    

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