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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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