<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
INF = 10**9
class SegTree:
    def __init__(self, ls):
        self.tree = [0]*(4 * len(ls) + 10)

        def build(t, l, r):
            if l == r:
                self.tree[t] = ls[l]
                return
            md = (l + r) // 2
            build(2 * t, l, md)
            build(2 * t + 1, md + 1, r)
            self.tree[t] = min(self.tree[2 * t], self.tree[2 * t + 1])
        build(1, 0, len(ls) - 1)
    
    def query(self, t, ql, qr, l, r):
        if ql <= l and r <= qr:
            return self.tree[t]
        if l > qr or r < ql:
            return INF
        md = (l + r) // 2
        return min(self.query(2 * t, ql, qr, l, md), self.query(2 * t + 1, ql, qr, md + 1, r))

def buildVis(root, n):
    h = []
    firstVis = [-1] * (n + 1)
    def dfs(node, dpth):
        firstVis[node.val] = len(h)
        h.append(dpth)
        if node.left != None:
            dfs(node.left, dpth + 1)
            h.append(dpth)
        if node.right != None:
            dfs(node.right, dpth + 1)
            h.append(dpth)
    dfs(root, 0)
    return h, firstVis

def getInvMap(root, out):
    if root == None:
        return
    out[root.val] = root
    getInvMap(root.left, out)
    getInvMap(root.right, out)

def buildPMap(root, out):
    if root == None:
        return
    if root.left != None:
        out[root.left] = root
        buildPMap(root.left, out)
    if root.right != None:
        out[root.right] = root
        buildPMap(root.right, out)

def climbTo(h, node, pMap):
    if h == 0:
        return []
    p = pMap[node]
    ls = climbTo(h - 1, p, pMap)
    if node == p.left:
        ls.append('L')
    else:
        ls.append('R')
    return ls
    

class Solution:
    def getDirections(self, root: Optional[TreeNode], startValue: int, destValue: int) -> str:
        invMap = {}
        getInvMap(root, invMap)
        pMap = {}
        buildPMap(root, pMap)
        n = len(invMap)
        h, firstVis = buildVis(root, n)
        s_tree = SegTree(h)
        rng = (firstVis[startValue], firstVis[destValue])
        mn_h = s_tree.query(1, min(rng), max(rng), 0, len(h) - 1)

        ordr = climbTo(h[firstVis[destValue]] - mn_h, invMap[destValue], pMap)
        return 'U'*(h[firstVis[startValue]] - mn_h) + ''.join(ordr)
78218