<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:
    void dfs_fa(TreeNode*root,TreeNode*father,map<int,TreeNode*>&fa,map<int,TreeNode*>&now)
    {
        if(root==nullptr)return;
        fa[root->val]=father;
        now[root->val]=root;
        dfs_fa(root->left,root,fa,now);
        dfs_fa(root->right,root,fa,now);
    }
    // string dfs_ans(TreeNode*root,int dest,map<int,TreeNode*>fa,vector<int>vis,string path)
    // {
    //     int now=root->val;
    //     if(now==dest)
    // }
    string getDirections(TreeNode* root, int startValue, int destValue) {
        map<int,TreeNode*>fa;
        map<int,TreeNode*>now;
        dfs_fa(root,nullptr,fa,now);
        string ans="";
        TreeNode* now_left=now[startValue];
        map<int,int>pre_sa;
        pre_sa[now_left->val]=0;
        int pre_cnt=0;
        while(now_left!=nullptr)
        {
            pre_cnt+=1;
            now_left=fa[now_left->val];
            if(now_left==nullptr)
            break;
            pre_sa[now_left->val]=pre_cnt;
        }
        string half_ans="";
        now_left=now[destValue];
        while(pre_sa.count(now_left->val)==0)
        {
           // cout<<"now left: "<<now_left->val<<'\n';
            auto now_left_fa=fa[now_left->val];
            if(now_left_fa->left!=nullptr&&now_left_fa->left->val==now_left->val)
            ans+='L';
            else
            ans+='R';
            now_left=now_left_fa;
        }
        reverse(ans.begin(),ans.end());
        string final_ans="";
        for(int i=0;i<pre_sa[now_left->val];i++)
        final_ans+='U';
        final_ans=final_ans+ans;
        return final_ans;
    //     queue<pair<int,string>>sa;
    //     sa.push({startValue,""});
    //     unordered_map<int,int>vis;
    //     //vis[startValue]=1;
    //    // cout<<1<<'\n';
    //     while(!sa.empty())
    //     {
    //         auto walk_now=sa.front();
    //         auto tree_now_node=now[walk_now.first];
    //         sa.pop();
    //         //cout<<"now dir: "<<walk_now.first<<" size: "<<sa.size()<<'\n';
    //         if(vis[walk_now.first])
    //         continue;
    //         vis[walk_now.first]=1;
    //         // if(sa.size()!=0)
          
    //         if(walk_now.first==destValue)
    //         return walk_now.second;
    //         auto now_father=fa[walk_now.first];
    //         // if(now_father==nullptr)
    //         // continue;
    //         if(now_father!=nullptr&&vis.count(now_father->val)==0)
    //         {
    //             //vis[now_father->val]=1;
    //             sa.push({now_father->val,walk_now.second+'U'});
    //         }
    //         if(tree_now_node->left!=nullptr&&vis.count(tree_now_node->left->val)==0)
    //         {
    //            // vis[tree_now_node->left->val]=1;
    //             sa.push({tree_now_node->left->val,walk_now.second+"L"});
    //         }
    //         if(tree_now_node->right!=nullptr&&vis.count(tree_now_node->right->val)==0)
    //         {
    //             //vis[tree_now_node->right->val]=1;
    //             sa.push({tree_now_node->right->val,walk_now.second+"R"});
    //         }
    //     }
    //     return "";
    }
};

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