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1372. Longest ZigZag Path in a Binary Tree

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

  • /**
 * 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.
# 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.
 * 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
}

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