1372. Longest ZigZag Path in a Binary Tree

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# 題目敘述

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.

# 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

# 解題思路

# Code

class Solution {

public:

    vector<int> dfs (TreeNode* root) {

        if (!root) return {-1, -1, -1};

        vector<int> leftSubtree = dfs(root->left);

        vector<int> rightSubtree = dfs(root->right);

        int leftlen = leftSubtree[1] + 1;

        int rightlen = rightSubtree[0] + 1;

        int mxlen = max({leftlen, rightlen, leftSubtree[2], rightSubtree[2]});

        return {leftlen, rightlen, mxlen};

    }

    int longestZigZag(TreeNode* root) {

        vector<int> res = dfs(root);

        return res[2];

    }

};

class Solution:

    def longestZigZag(self, root: Optional[TreeNode]) -> int:

        def dfs(root):

            if not root: return [-1, -1, -1]

            leftSubtree, rightSubtree = dfs(root.left), dfs(root.right)

            return [

                leftSubtree[1] + 1,

                rightSubtree[0] + 1,

                max(leftSubtree[1] + 1, rightSubtree[0] + 1, leftSubtree[2], rightSubtree[2])

            ]

        return dfs(root)[-1]