Class Solution
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- g1601_1700.s1609_even_odd_tree.Solution
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public class Solution extends Object
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, returntrue
if the binary tree is Even-Odd , otherwise returnfalse
.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
- The root of the binary tree is at level index
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Constructor Summary
Constructors Constructor Description Solution()
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Method Detail
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isEvenOddTree
public boolean isEvenOddTree(TreeNode root)
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