Question

1372. Longest ZigZag Path in a Binary Tree

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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

Solution 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.
 * 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.
 * 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
}