# 1372. Longest ZigZag Path in a Binary Tree

## Description

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]
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

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans;

public int longestZigZag(TreeNode root) {
dfs(root, 0, 0);
return ans;
}

private void dfs(TreeNode root, int l, int r) {
if (root == null) {
return;
}
ans = Math.max(ans, Math.max(l, r));
dfs(root.left, r + 1, 0);
dfs(root.right, 0, l + 1);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans = 0;

int longestZigZag(TreeNode* root) {
dfs(root, 0, 0);
return ans;
}

void dfs(TreeNode* root, int l, int r) {
if (!root) return;
ans = max(ans, max(l, r));
dfs(root->left, r + 1, 0);
dfs(root->right, 0, l + 1);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def longestZigZag(self, root: TreeNode) -> int:
def dfs(root, l, r):
if root is None:
return
nonlocal ans
ans = max(ans, l, r)
dfs(root.left, r + 1, 0)
dfs(root.right, 0, l + 1)

ans = 0
dfs(root, 0, 0)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func longestZigZag(root *TreeNode) int {
ans := 0
var dfs func(root *TreeNode, l, r int)
dfs = func(root *TreeNode, l, r int) {
if root == nil {
return
}
ans = max(ans, max(l, r))
dfs(root.Left, r+1, 0)
dfs(root.Right, 0, l+1)
}
dfs(root, 0, 0)
return ans
}