<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*> parents;
    unordered_map<TreeNode*, pair<char, TreeNode*>> prev;
    TreeNode* start_node;
    string getDirections(TreeNode* root, int start, int dest) {
        explore(root, nullptr, start);
        // return "";
        return shortest_path(start_node, dest);
    }
    
    void explore(TreeNode* node, TreeNode* parent, int start) {
        if(node == nullptr) return;
        if(node->val == start) start_node = node;
        parents[node->val] = parent;
        explore(node->left, node, start);
        explore(node->right, node, start);
    }
    
    string shortest_path(TreeNode* n, int target_value) {
        unordered_set<TreeNode*> seen;
        queue<TreeNode*> q;
        q.push(n);
        seen.insert(n);
        while(!q.empty()) {
            auto node = q.front(); q.pop();
            if(node->val == target_value) {
                n = node;
                string res;
                while(prev.contains(n)) {
                    res += prev[n].first;
                    n = prev[n].second;
                }
                reverse(res.begin(), res.end());
                return res;
            }
            
            if(parents[node->val] && seen.find(parents[node->val]) == seen.end()) {
                q.push(parents[node->val]);
                seen.insert(parents[node->val]);
                prev[parents[node->val]] = {'U', node};
            }
            if(node->left && seen.find(node->left) == seen.end()) {
                q.push(node->left);
                seen.insert(node->left);
                prev[node->left] = {'L', node};
            }
            if(node->right && seen.find(node->right) == seen.end()) {
                q.push(node->right);
                seen.insert(node->right);
                prev[node->right] = {'R', node};
            }
        }
        return "";
        
    }
};
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