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

<p>Find the <strong>shortest path</strong> starting from node <code>s</code> and ending at node <code>t</code>. Generate step-by-step directions of such path as a string consisting of only the <strong>uppercase</strong> letters <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, and <code>&#39;U&#39;</code>. Each letter indicates a specific direction:</p>

<ul>
	<li><code>&#39;L&#39;</code> means to go from a node to its <strong>left child</strong> node.</li>
	<li><code>&#39;R&#39;</code> means to go from a node to its <strong>right child</strong> node.</li>
	<li><code>&#39;U&#39;</code> means to go from a node to its <strong>parent</strong> node.</li>
</ul>

<p>Return <em>the step-by-step directions of the <strong>shortest path</strong> from node </em><code>s</code><em> to node</em> <code>t</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/15/eg1.png" style="width: 214px; height: 163px;" />
<pre>
<strong>Input:</strong> root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6
<strong>Output:</strong> &quot;UURL&quot;
<strong>Explanation:</strong> The shortest path is: 3 &rarr; 1 &rarr; 5 &rarr; 2 &rarr; 6.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/15/eg2.png" style="width: 74px; height: 102px;" />
<pre>
<strong>Input:</strong> root = [2,1], startValue = 2, destValue = 1
<strong>Output:</strong> &quot;L&quot;
<strong>Explanation:</strong> The shortest path is: 2 &rarr; 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is <code>n</code>.</li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= Node.val &lt;= n</code></li>
	<li>All the values in the tree are <strong>unique</strong>.</li>
	<li><code>1 &lt;= startValue, destValue &lt;= n</code></li>
	<li><code>startValue != destValue</code></li>
</ul>

/**
 * 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:
    unordered_map<int, TreeNode*>parent;
    void gp(TreeNode* root, TreeNode* par) {
        if(root == NULL) return;

        parent[root->val] = par;
        gp(root->left, root);
        gp(root->right, root);
        return;
    }

    TreeNode* findBy(TreeNode* root, int startValue) {
        if(root == NULL) return NULL;

        if(root->val == startValue) return root;

        TreeNode* l = findBy(root->left, startValue);

        if(l != NULL) return l;

        return findBy(root->right, startValue);
    }

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

        TreeNode* start = findBy(root, startValue);

        queue<TreeNode*>bfs;
        map < TreeNode*, bool> visited;
        map < TreeNode*, TreeNode* > dir;

        bfs.push(start);

        while(!bfs.empty()) {
            auto curr = bfs.front();
            bfs.pop();
            TreeNode* currNode = curr;

            if(visited[currNode]) continue;
            visited[currNode] = true;

            if(currNode == NULL) continue;

            if(currNode->val == destValue)  {
                auto dst = curr;
                vector < TreeNode* > path = {dst};


                while(dst != start) {
                    path.push_back(dir[dst]);
                    dst = dir[dst];
                }

                // path.push_back(start);
                reverse(path.begin(), path.end());

                string ans = "";
                for(int i=0; i<path.size()-1; i++) {
                    if(path[i+1] == parent[path[i]->val]) ans += 'U';
                    if(path[i+1] == path[i]->left) ans += 'L';
                    if(path[i+1] == path[i]->right) ans += 'R';
                }

                return ans;
            }

            if(parent[currNode->val] != NULL) {
                bfs.push(parent[currNode->val]);
                if(dir.find(parent[currNode->val]) == dir.end()) {
                    dir[parent[currNode->val]] = currNode;
                }
            }

            if(currNode->left != NULL) {
                bfs.push(currNode->left);
                if(dir.find(currNode->left) == dir.end()) {
                    dir[currNode->left] = currNode;
                }
            }

            if(currNode->right != NULL) {
                bfs.push(currNode->right);
                if(dir.find(currNode->right) == dir.end()) {
                    dir[currNode->right] = currNode;
                }
            }
        }

        return "";
    }
};
222688