<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* st=0;     map<int,int>mp1;
    string ans="";
    string getDirections(TreeNode* root, int startValue, int destValue) {
        map<TreeNode*,TreeNode*>mp;
   
        fun(mp,root,startValue);
        mp[root]=0;
        string s="";
        find(st,mp,destValue,s);
        return ans;
    }
    void find(TreeNode* t,map<TreeNode*,TreeNode*>&mp,int d,string &s){
        if(t==0)return;
        if(t->val==d){
            if(ans.length()==0)ans=s;
            else if(s.length()!=0){
                if(s.length()<=ans.length()){
                    ans=s;
                }
            }
            //cout<<ans<<" ";
        }
        mp1[t->val]=1;
        if(t->left&&mp1[t->left->val]==0){
            s+="L";
            mp1[t->left->val]=1;
           find(t->left,mp,d,s);
            mp1[t->left->val]=0;
            s.pop_back();
        }
        if(t->right&& mp1[t->right->val]==0){
            s+="R";
             mp1[t->right->val]=1;
            find(t->right,mp,d,s);
             mp1[t->right->val]=0;
              s.pop_back();
        }
        if(mp[t]!=0&& mp1[mp[t]->val]==0){
            s+="U";
             mp1[mp[t]->val]=1;
            find(mp[t],mp,d,s);
             mp1[mp[t]->val]=0;
              s.pop_back();
        }
    }
    void fun(map<TreeNode*,TreeNode*>&mp,TreeNode* t,int start){
        if(t==0)return;
        mp1[t->val]=0;
        if(t->val==start){
            st=t;
        }
        if(t->left){
            mp[t->left]=t;
        }
        if(t->right){
            mp[t->right]=t;
        }
        fun(mp,t->left,start);
        fun(mp,t->right,start);
    }
};
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