LeetCode in Kotlin
1372. Longest ZigZag Path in a Binary Tree
Medium
You are given the root of a binary tree.
A ZigZag path for a binary tree is defined as follow:
- Choose any node in the binary tree and a direction (right or left).
- If the current direction is right, move to the right child of the current node; otherwise, move to the left child.
- Change the direction from right to left or from left to right.
- Repeat the second and third steps until you can’t move in the tree.
Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of 0).
Return the longest ZigZag path contained in that tree.
Example 1:
Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1,null,1]
Output: 3
Explanation: Longest ZigZag path in blue nodes (right -> left -> right).
Example 2:
Input: root = [1,1,1,null,1,null,null,1,1,null,1]
Output: 4
Explanation: Longest ZigZag path in blue nodes (left -> right -> left -> right).
Example 3:
Input: root = [1]
Output: 0
Constraints:
- The number of nodes in the tree is in the range
[1, 5 * 104]. 1 <= Node.val <= 100
Solution
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
private var maxLength = 0
fun longestZigZag(root: TreeNode?): Int {
dfs(root, true)
return maxLength
}
private fun dfs(root: TreeNode?, isLeft: Boolean): Int {
if (root == null) {
return 0
}
val left = dfs(root.left, false)
val right = dfs(root.right, true)
maxLength = Math.max(maxLength, left)
maxLength = Math.max(maxLength, right)
return 1 + if (isLeft) left else right
}
}