Leet Code Daily 2024-07-16 - Step-By-Step Directions From a Binary Tree Node to Another

/**
 * 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:
    void dfs(TreeNode* root, unordered_map<int, TreeNode*>& val_to_ptr,
             unordered_map<TreeNode*, TreeNode*>& parent) {
        if (root != nullptr) {
            val_to_ptr[root->val] = root;
            if (root->left != nullptr) {
                parent[root->left] = root;
                dfs(root->left, val_to_ptr, parent);
            }
            if (root->right != nullptr) {
                parent[root->right] = root;
                dfs(root->right, val_to_ptr, parent);
            }
        }
    }

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

        unordered_map<int, TreeNode*> val_to_ptr;
        unordered_map<TreeNode*, TreeNode*> parent;
        dfs(root, val_to_ptr, parent);

        unordered_set<TreeNode*> dest_tree_path;
        for (auto ptr = val_to_ptr[destValue]; ptr != nullptr;
             ptr = parent[ptr]) {
            dest_tree_path.insert(ptr);
        }

        string ret = "";
        for (auto ptr = val_to_ptr[startValue]; ptr != val_to_ptr[destValue];) {
            if (ptr->left != nullptr && dest_tree_path.contains(ptr->left)) {
                ret.push_back('L');
                ptr = ptr->left;
                continue;
            }
            if (ptr->right != nullptr && dest_tree_path.contains(ptr->right)) {
                ret.push_back('R');
                ptr = ptr->right;
                continue;
            }
            ptr = parent[ptr];
            ret.push_back('U');
        }
        return ret;
    }
};