1300 - 1399
1372 - Longest ZigZag Path in a Binary Tree (Medium)
Problem Link
https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree/
Problem Statement
You are given the root of a binary tree.
A ZigZag path for a binary tree is defined as follow:
- Choose anynode 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: 0Constraints:
- The number of nodes in the tree is in the range
[1, 5 * 10^4]. 1 <= Node.val <= 100
Approach 1: DFS
/**
* 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 dfs(TreeNode* root, int isLeft, int cnt) {
// root is nullptr, we can return `cnt`
if (!root) return cnt;
// if `isLeft` is true, we have two choices
// 1. go to right making a zipzag path - increase the cnt by 1
// 2. still go to left - starting a new zigzag path - hence cnt is set to 0
if (isLeft) return max(dfs(root->right, 0, cnt + 1), dfs(root->left, 1, 0));
// similarly, we apply the same logic for the opposite direction
return max(dfs(root->left, 1, cnt + 1), dfs(root->right, 0, 0));
}
int longestZigZag(TreeNode* root) {
// go through left-subtree and right-subtree
// get the max result
return max(dfs(root->left, 1, 0), dfs(root->right, 0, 0));
}
};# 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: Optional[TreeNode]) -> int:
def dfs(node, isLeft, cnt):
# root is None, we can return `cnt`
if node is None: return cnt
# if `isLeft` is true, we have two choices
# 1. go to right making a zipzag path - increase the cnt by 1
# 2. still go to left - starting a new zigzag path - hence cnt is set to 0
if isLeft: return max(dfs(node.right, 0, cnt + 1), dfs(node.left, 1, 0))
# similarly, we apply the same logic for the opposite direction
return max(dfs(node.left, 1, cnt + 1), dfs(node.right, 0, 0))
# go through left-subtree and right-subtree and get the max result
return max(dfs(root.left, 1, 0), dfs(root.right, 0, 0))