1372.Longest ZigZag Path in a Binary Tree

tags: Medium,Tree,DP,DFS

1372. 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 * 10⁴].
• 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;
}

MarsgoatApr 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

Ron ChenWed, Apr 19, 2023

Reference

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