<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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getDirections(self, root: Optional[TreeNode], startValue: int, destValue: int) -> str:
        def lca(cur) -> Optional[TreeNode]:
            if not cur.left and not cur.right:
                if cur.val == startValue or cur.val == destValue:
                    return cur 
                return None 
            left = None 
            if cur.left:
                left = lca(cur.left)
            right = None 
            if cur.right:
                right = lca(cur.right)
            # cur is lca
            if left and right:
                return cur 
            elif left:
                if left.val == startValue and cur.val == destValue:
                    return cur 
                if left.val == destValue and cur.val == startValue:    
                    return cur 
                return left 
            elif right:
                if right.val == startValue and cur.val == destValue:
                    return cur 
                if right.val == destValue and cur.val == startValue:
                    return cur 
                return right 
            else:
                if cur.val == startValue or cur.val == destValue:
                    return cur
                return None 
        my_lca = lca(root)
        
        
        parents = {my_lca.val : (None, None)}
        def dfs(cur):
            if not cur: return 
            if cur.left:
                parents[cur.left.val] = (cur.val, "L")
            dfs(cur.left)
            if cur.right:
                parents[cur.right.val] = (cur.val, "R")
            dfs(cur.right)
        dfs(my_lca)
        
        
        first_str, second_str = "", ""
        start = startValue
        end = destValue
        while True:
            parent, direction = parents[start]
            if not parent:
                break
            first_str += "U"
            start = parent
        while True:
            parent, direction = parents[end]
            if not parent:
                break
            second_str = direction + second_str
            end = parent
        return first_str + second_str
        
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