<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 {
    void dfs(TreeNode* node, int target, stack<string> &dir, bool &done){
        if (done || node == NULL) return;
        if (node -> val == target){
            done = true;
            return;
        }
        if (done || node == NULL) return;
        dir.push("L");
        dfs(node -> left, target, dir, done);
        if (done) return;
        dir.pop();
        dir.push("R");
        dfs(node -> right, target, dir, done);
        if (done) return;
        dir.pop();
    }
public:
    string getDirections(TreeNode* root, int startValue, int destValue) {
        bool done1 = false, done2 = false;
        stack<string> dir1;
        stack<string> dir2;
        dfs(root, startValue, dir1, done1);
        dfs(root, destValue, dir2, done2);
        string dirxn1 = "", dirxn2 = "", ans = "";
        dirxn1 += "x";
        dirxn2 += "x";
        while(!dir1.empty()){
            dirxn1 += dir1.top();
            dir1.pop();
        }
        while(!dir2.empty()){
            dirxn2 += dir2.top();
            dir2.pop();
        }
        cout<<dirxn1<<' '<<dirxn2<<'\n';
        bool start = false;
        int i = dirxn1.length()-1;
        int j = dirxn2.length()-1;
        cout<<i<<' '<<j<<'\n';
        int numU = 0;
        while(i > 0 && j > 0){
            cout<<start<<' '<<dirxn1[i]<<' '<<dirxn2[j]<<'\n';
            if (!start && dirxn1[i] != dirxn2[j]) {
                start = true;
                numU = i;
            }
            if (start) ans += dirxn2[j];
            i--;
            j--;
        }
        while(j > 0) {
            ans += dirxn2[j];
            j--;
        }
        if (i>j && numU == 0) numU = i-j;
        reverse(ans.begin(), ans.end());
        for (int i=0; i<numU; i++) ans += "U";
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
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