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
}