<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>

class Solution {
public:
    string getDirections(TreeNode* root, int startValue, int destValue) {
        // Map to store parent nodes for quick upward traversal
        unordered_map<int, TreeNode*> parentMap;
        
        // Find the start node and populate parent map in a single traversal
        TreeNode* startNode = findStartNode(root, startValue);
        populateParentMap(root, parentMap);
        
        // Perform BFS to find the path from start to destination
        queue<TreeNode*> q;
        q.push(startNode);
        unordered_set<TreeNode*> visitedNodes;
        // Key: next node, Value: <current node, direction>
        unordered_map<TreeNode*, pair<TreeNode*, string>> pathTracker;
        visitedNodes.insert(startNode);
        
        while (!q.empty()) {
            TreeNode* currentNode = q.front();
            q.pop();
            
            // If destination is reached, return the path
            if (currentNode->val == destValue) {
                return backtrackPath(currentNode, pathTracker);
            }
            
            // Check and add parent node (upward direction)
            if (parentMap.find(currentNode->val) != parentMap.end()) {
                TreeNode* parentNode = parentMap[currentNode->val];
                if (visitedNodes.find(parentNode) == visitedNodes.end()) {
                    q.push(parentNode);
                    pathTracker[parentNode] = {currentNode, "U"};
                    visitedNodes.insert(parentNode);
                }
            }
            
            // Check and add left child
            if (currentNode->left && visitedNodes.find(currentNode->left) == visitedNodes.end()) {
                q.push(currentNode->left);
                pathTracker[currentNode->left] = {currentNode, "L"};
                visitedNodes.insert(currentNode->left);
            }
            
            // Check and add right child
            if (currentNode->right && visitedNodes.find(currentNode->right) == visitedNodes.end()) {
                q.push(currentNode->right);
                pathTracker[currentNode->right] = {currentNode, "R"};
                visitedNodes.insert(currentNode->right);
            }
        }
        
        // This line should never be reached if the tree is valid and contains both nodes
        return "";
    }

private:
    // Backtrack the path from destination to start and reverse it
    string backtrackPath(TreeNode* node, unordered_map<TreeNode*, pair<TreeNode*, string>>& pathTracker) {
        string path;
        // Construct the path in reverse order
        while (pathTracker.count(node)) {
            path += pathTracker[node].second;
            node = pathTracker[node].first;
        }
        // Reverse the path to get the correct order from start to destination
        reverse(path.begin(), path.end());
        return path;
    }

    // Populate the parent map for quick upward traversal
    void populateParentMap(TreeNode* node, unordered_map<int, TreeNode*>& parentMap) {
        if (!node) return;
        // Add children to the map and recurse further
        if (node->left) {
            parentMap[node->left->val] = node;
            populateParentMap(node->left, parentMap);
        }
        if (node->right) {
            parentMap[node->right->val] = node;
            populateParentMap(node->right, parentMap);
        }
    }

    // Find the start node in the tree
    TreeNode* findStartNode(TreeNode* node, int startValue) {
        if (!node) return nullptr;
        if (node->val == startValue) return node;
        TreeNode* leftResult = findStartNode(node->left, startValue);
        // If left subtree returns a node, it must be the start node. Return it.
        // Otherwise, return whatever is returned by the right subtree.
        return leftResult ? leftResult : findStartNode(node->right, startValue);
    }
};
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