<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 _find_start_node(self, node, start_value):
        if not node:
            return None
        if node.val == start_value:
            return node
        left_result = self._find_start_node(node.left, start_value)
        
        if left_result:
            return left_result
        
        # otherwise, it is found on the right subtree.
        return self._find_start_node(node.right, start_value)

    def _populate_parent_map(self, node, parent_map):
        if not node:
            return
        
        # add children to the map & recurse further
        if node.left:
            parent_map[node.left.val] = node
            self._populate_parent_map(node.left, parent_map)
        
        if node.right:
            parent_map[node.right.val] = node
            self._populate_parent_map(node.right, parent_map)

    def getDirections(self, root: Optional[TreeNode], startValue: int, destValue: int) -> str:
        # map to store parent nodes
        parent_map = {}

        # find the start node and populate parent map
        start_node = self._find_start_node(root, startValue)
        self._populate_parent_map(root, parent_map)

        # BFS to find the path
        queue = deque([start_node])
        visited_nodes = set()
        # key: next_node, value: <curr_node, direction>
        path_tracker = {}
        visited_nodes.add(start_node)

        while queue:
            curr_node = queue.popleft()

            # if dest is reached, return the path
            if curr_node.val == destValue:
                return self._backtrack_path(curr_node, path_tracker)
            
            # check & add parent node
            if curr_node.val in parent_map:
                parent_node = parent_map[curr_node.val]
                if parent_node not in visited_nodes:
                    queue.append(parent_node)
                    path_tracker[parent_node] = (curr_node, 'U')
                    visited_nodes.add(parent_node)
            
            # check & add left child node
            if curr_node.left and curr_node.left not in visited_nodes:
                queue.append(curr_node.left)
                path_tracker[curr_node.left] = (curr_node, "L")
                visited_nodes.add(curr_node.left)
            
            # check & add right child node
            if curr_node.right and curr_node.right not in visited_nodes:
                queue.append(curr_node.right)
                path_tracker[curr_node.right] = (curr_node, "R")
                visited_nodes.add(curr_node.right)
            
        # This should never be reached if tree is valid
        return ""
        
    def _backtrack_path(self, node, path_tracker):
        path = []
        # construct the path:
        while node in path_tracker:
            # add direction in reverse order and move to prev node
            path.append(path_tracker[node][1]) # 'U', 'L', 'R'
            node = path_tracker[node][0]
        
        path.reverse()
        return "".join(path)





        
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