<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

# Anh
class Solution:
    def getDirections(self, root: Optional[TreeNode], startValue: int, destValue: int) -> str:
        # 1st approach: convert into a graph, where the edges contain the directions. Do a BFS from start to dest
        # 2nd approach: DFS for start & dest separately, collect their paths, then eliminate commonalities
        # solution: 1st is good! 2nd: best.

        # root is never optional, since n >= 2

        startValuePath = []
        destValuePath = []
        foundNodes = 0
        
        def dfs(node: TreeNode, path: List[int]) -> bool:
            nonlocal startValue
            nonlocal destValue
            nonlocal startValuePath
            nonlocal destValuePath
            nonlocal foundNodes

            # returns whether target is found
            if node.val == startValue:
                startValuePath = path.copy()
                foundNodes += 1
                if foundNodes == 2:
                    return True
            elif node.val == destValue:
                destValuePath = path.copy()
                foundNodes += 1
                if foundNodes == 2:
                    return True

            leftResult = False
            if node.left:
                path.append("L")
                leftResult = dfs(node.left, path)
                if not leftResult:
                    path.pop()
                else:
                    return True

            rightResult = False
            if node.right:
                path.append("R")
                rightResult = dfs(node.right, path)
                if not rightResult:
                    path.pop()
                else:
                    return True
            
            return False
        
        path = []
        dfs(root, path)
        
        s = len(startValuePath)
        d = len(destValuePath)
        sToDPath = ""
        # special case
        if s == 0: # start == root
            dPath = ""
            for move in destValuePath:
                dPath += move
            return dPath
        elif d == 0: # d
            sPath = "U" * len(startValuePath)
            return sPath

        i = 0
        while i < s and i < d and startValuePath[i] == destValuePath[i]:
            i += 1
        
        lcaInd = i # lowest common ancestor
        while i < s:
            move = startValuePath[i]
            if move == "L" or move == "R":
                move = "U"
            sToDPath += move
            i += 1
        i = lcaInd
        while i < d:
            sToDPath += destValuePath[i]
            i += 1
        
        return sToDPath
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