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
Solution 1: DFS
Time complexity: O(n)
Space complexity: O(n)
C++
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class Solution { public: bool isEvenOddTree(TreeNode* root) { vector<int> vals; function<bool(TreeNode*, int)> dfs = [&](TreeNode* root, int d) { if (!root) return true; if (vals.size() <= d) vals.push_back(d % 2 == 0 ? -1 : INT_MAX); int& val = vals[d]; if (d % 2 == 0) if (root->val % 2 == 0 || root->val <= val) return false; if (d % 2 == 1) if (root->val % 2 == 1 || root->val >= val) return false; val = root->val; return dfs(root->left, d + 1) && dfs(root->right, d + 1); }; return dfs(root, 0); } }; |
Python3
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# Author: Huahua class Solution: def isEvenOddTree(self, root: TreeNode) -> bool: vals = {} def dfs(root: TreeNode, d: int) -> bool: if not root: return True if d not in vals: vals[d] = 0 if d % 2 == 0 else 10**7 comp = int.__ge__ if d % 2 else int.__le__ if root.val % 2 == d % 2 or comp(root.val, vals[d]): return False vals[d] = root.val return dfs(root.left, d + 1) and dfs(root.right, d + 1) return dfs(root, 0) |
Solution 2: BFS
Time complexity: O(n)
Space complexity: O(n)
Python3
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# Author: Huahua class Solution: def isEvenOddTree(self, root: TreeNode) -> bool: q = collections.deque([root]) odd = 1 while q: prev = 0 if odd else 10**7 for _ in range(len(q)): root = q.popleft() if not root: continue comp = int.__le__ if odd else int.__ge__ if root.val % 2 != odd or comp(root.val, prev): return False prev = root.val q += (root.left, root.right) odd = 1 - odd return True |
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