2096. Step-By-Step Directions From a Binary Tree Node to Another | LeetCode Solution

 2096. Step-By-Step Directions From a Binary Tree Node to Another | LeetCode Solution

You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value from 1 to n. You are also given an integer startValue representing the value of the start node s, and a different integer destValue representing the value of the destination node t.

Find the shortest path starting from node s and ending at node t. Generate step-by-step directions of such path as a string consisting of only the uppercase letters 'L', 'R', and 'U'. Each letter indicates a specific direction:

• 'L' means to go from a node to its left child node.
• 'R' means to go from a node to its right child node.
• 'U' means to go from a node to its parent node.

Return the step-by-step directions of the shortest path from node s to node t.

 

Example 1:

Input: root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6
Output: "UURL"
Explanation: The shortest path is: 3 → 1 → 5 → 2 → 6.

Example 2:

Input: root = [2,1], startValue = 2, destValue = 1
Output: "L"
Explanation: The shortest path is: 2 → 1.

 

Constraints:

• The number of nodes in the tree is n.
• 2 <= n <= 105
• 1 <= Node.val <= n
• All the values in the tree are unique.
• 1 <= startValue, destValue <= n
• startValue != destValue

/**

 * 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:

  bool find(TreeNode *root, int dest, string &path)

  {

    if (root->val == dest)

      return true;

    if (root->left && find(root->left, dest, path))

      path.push_back('L');

    else if (root->right && find(root->right, dest, path))

      path.push_back('R');

    return !path.empty();

  }

  string getDirections(TreeNode *root, int startValue, int destValue)

  {

    string one, two;

    find(root, startValue, one);

    find(root, destValue, two);

    while (!one.empty() && !two.empty() && one.back() == two.back())

    {

      one.pop_back();

      two.pop_back();

    }

    reverse(two.begin(), two.end());

    return string(one.size(), 'U') + two;

  }

};