1372.Longest ZigZag Path in a Binary Tree
===

###### tags: `Medium`,`Tree`,`DP`,`DFS`

[1372. Longest ZigZag Path in a Binary Tree](https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree/)

### 題目描述

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.*

### 範例

**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

### 解答
#### Javascript

function longestZigZag(root) {
  let longest = 0;
  const stack = [[root, 0, 0]];
  while (stack.length) {
    const [node, left, right] = stack.pop();
    if (node === null) continue;
    longest = Math.max(longest, left, right);
    stack.push([node.left, right + 1, 0]);
    stack.push([node.right, 0, left + 1]);
  }

  return longest;
}

> [name=Marsgoat][time=Apr 19, 2023]

#### Python

class Solution:
    def longestZigZag(self, root: Optional[TreeNode]) -> int:
        self.pathLength = 0

        # goLeft: Next step go left or not
        def dfs(node, goLeft, steps):
            if node:
                self.pathLength = max(self.pathLength, steps)

                if goLeft:
                    dfs(node.left, False, steps + 1)  # Continue to the left
                    dfs(node.right, True, 1)  # Restart to the right
                else:
                    dfs(node.left, False, 1)  # Restart to the left
                    dfs(node.right, True, steps + 1)  # Continue to the right

        dfs(root, False, 0)  # Left start
        dfs(root, True, 0)  # Right start
        return self.pathLength

> [name=Ron Chen][time=Wed, Apr 19, 2023]

### Reference

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