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

https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree/

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init() { self.val = 0; self.left = nil; self.right = nil; }
 *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
 *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
 *         self.val = val
 *         self.left = left
 *         self.right = right
 *     }
 * }
 */
class Solution {
    func longestZigZag(_ root: TreeNode?) -> Int {
        dfs(root, true)
        return result
    }
    var result = 0
    func dfs(_ root: TreeNode?, _ isLeft: Bool) -> Int {
        if root == nil {
            return 0
        }
        var l = dfs(root!.left, false)
        var r = dfs(root!.right, true)
        result = max(result, l)
        result = max(result, r)
        if isLeft == true {
            return 1+l
        } else {
            return 1+r
        }
    }
}