<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:
    TreeNode* getPath(TreeNode* root, int &n, vector<pair<char,int>> &path, int l, int r,int* lmatch,int* rmatch)
    {
        if(!root){ *lmatch = 0; *rmatch=0; return NULL;}
        if(root->val == n)
        {
            //cout<<"MATCH";
            if(l) *lmatch = 1;
            if(r) *rmatch = 1;
            //cout<<"l"<<l<<" r"<<r;
            return root;
        }

        TreeNode* ret_left = getPath(root->left, n, path, 1,0,lmatch,rmatch);
        //cout<<"lmatch="<<*lmatch<<endl;
        if(*lmatch && ret_left) { 
            //cout<<"L";
            path.push_back(make_pair('L',ret_left->val)); 
            if(l) *lmatch = 1;
            if(r) *rmatch = 1;
            return root;
        }
        TreeNode* ret_right = getPath(root->right, n, path, 0,1,lmatch, rmatch);
        if(*rmatch && ret_right)
        {
            //cout<<"R";
            path.push_back(make_pair('R',ret_right->val));
            if(l) *lmatch = 1;
            if(r) *rmatch = 1;
            return root;
        }
        
        *lmatch = 0; *rmatch=0;
        return root;

    }
    TreeNode* LCA(TreeNode* root, int &n1, int &n2)
    {
        if(!root || root->val == n1 || root->val == n2) return root;
        TreeNode* left_node = LCA(root->left, n1, n2);
        TreeNode* right_node = LCA(root->right, n1, n2);
        if(left_node && right_node) return root;
        else return left_node ? left_node: right_node;
    }
    string getDirections(TreeNode* root, int startValue, int destValue) {

        TreeNode* lca = LCA(root, startValue, destValue);
        //cout<<lca->val<<". ";
        vector<pair<char,int>> path1, path2;
        int lmatch=0, rmatch=0;
        getPath(lca, startValue, path1, 1, 0, &lmatch, &rmatch);
        string result;
        //cout<<"path";
        for(auto it: path1)
        {
            //cout<<it.first<<it.second<<" ";
            result += string("U");
        }
        //cout<<endl;
        getPath(lca, destValue, path2, 0, 1, &lmatch, &rmatch);
        reverse(path2.begin(), path2.end());
        for(auto it: path2)
        {
            //cout<<it.first<<it.second<<" ";
            result += it.first;
        }
        return result;
        
    }
};
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