Formatted question description: https://leetcode.ca/all/1372.html

1372. Longest ZigZag Path in a Binary Tree (Medium)

Given a binary tree root, 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 then move to the right child of the current node otherwise move to the left child.
  • Change the direction from right to left or right to left.
  • Repeat the second and third step 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: 0

 

Constraints:

  • Each tree has at most 50000 nodes..
  • Each node's value is between [1, 100].

Related Topics:
Dynamic Programming, Tree

Solution 1.

// OJ: https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree
// Time: O(N)
// Space: O(H)
class Solution {
    int A = 0;
    pair<int, int> postorder(TreeNode* root)  {
        if (!root || (!root->left && !root->right)) return { 0, 0 };
        pair<int, int> ans;
        if (root->left) {
            auto left = postorder(root->left);
            ans.first = 1 + left.second;
        }
        if (root->right) {
            auto right = postorder(root->right);
            ans.second = 1 + right.first;
        }
        A = max({ A, ans.first, ans.second });
        return ans;
    }
public:
    int longestZigZag(TreeNode* root) {
        postorder(root);
        return A;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public int longestZigZag(TreeNode root) {
            Map<TreeNode, int[]> map = new HashMap<TreeNode, int[]>();
            map.put(root, new int[]{0, 0});
            int maxLength = 0;
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                TreeNode node = queue.poll();
                int[] lengths = map.get(node);
                TreeNode left = node.left, right = node.right;
                if (left != null) {
                    int[] leftLengths = new int[2];
                    leftLengths[0] = lengths[1] + 1;
                    maxLength = Math.max(maxLength, leftLengths[0]);
                    map.put(left, leftLengths);
                    queue.offer(left);
                }
                if (right != null) {
                    int[] rightLengths = new int[2];
                    rightLengths[1] = lengths[0] + 1;
                    maxLength = Math.max(maxLength, rightLengths[1]);
                    map.put(right, rightLengths);
                    queue.offer(right);
                }
            }
            return maxLength;
        }
    }
    
  • // OJ: https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree
    // Time: O(N)
    // Space: O(H)
    class Solution {
        int A = 0;
        pair<int, int> postorder(TreeNode* root)  {
            if (!root || (!root->left && !root->right)) return { 0, 0 };
            pair<int, int> ans;
            if (root->left) {
                auto left = postorder(root->left);
                ans.first = 1 + left.second;
            }
            if (root->right) {
                auto right = postorder(root->right);
                ans.second = 1 + right.first;
            }
            A = max({ A, ans.first, ans.second });
            return ans;
        }
    public:
        int longestZigZag(TreeNode* root) {
            postorder(root);
            return A;
        }
    };
    
  • print("Todo!")
    

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