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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 index1, their children are at level index2, 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; } }; -
# 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