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 * 10^4]
.1 <= Node.val <= 100
解題
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func longestZigZag(root *TreeNode) int {
ans := 0
var helper func(*TreeNode, int, bool)
helper = func(root *TreeNode, length int, goLeft bool) {
if root == nil { return }
ans = max(length, ans)
if goLeft {
helper(root.Left, length + 1, false) // 前面的長度加上這一條
helper(root.Right, 1, true) // 應該要往左走卻往右走,步數歸零在+1
} else {
helper(root.Right, length + 1, true)
helper(root.Left, 1, false)
}
}
helper(root, 0, true)
helper(root, 0, false)
return ans
}
func max(a, b int) int {
if a > b { return a }
return b
}
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