<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) {}
 * };
 */


TreeNode* source;

void dfs(TreeNode* root,TreeNode* parent,map<int,TreeNode*> &par,int s){
    if(!root){
        return ;
    }
    
    if(root->val==s){
        source=root;
    }
    
    if(parent){
        par[root->val]=parent;
    }
    dfs(root->left,root,par,s);
    dfs(root->right,root,par,s);
}

class Solution {
public:
    string getDirections(TreeNode* root, int s, int d) {
        
        
        map<int,TreeNode*> par;
        queue<TreeNode*> q;
        dfs(root,NULL,par,s);

        q.push(source);
        map<int,int> vis;
        map<int,pair<int,int>> p;
        vis[source->val]=1;
        
        while(!q.empty()){
            TreeNode* x = q.front();
            q.pop();
            if(x->val==d){
                 break;
            }
            
            if(x->left and !vis.count(x->left->val)){
                q.push(x->left);
                p[x->left->val]={x->val,0};
                vis[x->left->val]=1;
            }
            
            if(x->right and !vis.count(x->right->val)){
                q.push(x->right);
                p[x->right->val]={x->val,1};
                vis[x->right->val]=1;
            }
            
            if(par.count(x->val) and !vis.count(par[x->val]->val)){
                q.push(par[x->val]);
                p[par[x->val]->val]={x->val,-1};
                vis[par[x->val]->val]=1;
            }
        }
        
        string res="";
        int end=d;
        while(end!=s){
            pair<int,int> x=p[end];
            if(x.second==0){
                res+="L";
            }
            else if(x.second==1){
                res+="R";
            }
            else{
                res+="U";
            }
            end=p[end].first;
        }
        
        reverse(res.begin(),res.end());
        return res;
    }
};
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