<p>There is an undirected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given the integer <code>n</code> and a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with weight <code>w<sub>i</sub></code> in the tree.</p>

<p>You are also given a 2D integer array <code>queries</code> of length <code>m</code>, where <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code>. For each query, find the <strong>minimum number of operations</strong> required to make the weight of every edge on the path from <code>a<sub>i</sub></code> to <code>b<sub>i</sub></code> equal. In one operation, you can choose any edge of the tree and change its weight to any value.</p>

<p><strong>Note</strong> that:</p>

<ul>
	<li>Queries are <strong>independent</strong> of each other, meaning that the tree returns to its <strong>initial state</strong> on each new query.</li>
	<li>The path from <code>a<sub>i</sub></code> to <code>b<sub>i</sub></code> is a sequence of <strong>distinct</strong> nodes starting with node <code>a<sub>i</sub></code> and ending with node <code>b<sub>i</sub></code> such that every two adjacent nodes in the sequence share an edge in the tree.</li>
</ul>

<p>Return <em>an array </em><code>answer</code><em> of length </em><code>m</code><em> where</em> <code>answer[i]</code> <em>is the answer to the</em> <code>i<sup>th</sup></code> <em>query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/11/graph-6-1.png" style="width: 339px; height: 344px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[0,1,1],[1,2,1],[2,3,1],[3,4,2],[4,5,2],[5,6,2]], queries = [[0,3],[3,6],[2,6],[0,6]]
<strong>Output:</strong> [0,0,1,3]
<strong>Explanation:</strong> In the first query, all the edges in the path from 0 to 3 have a weight of 1. Hence, the answer is 0.
In the second query, all the edges in the path from 3 to 6 have a weight of 2. Hence, the answer is 0.
In the third query, we change the weight of edge [2,3] to 2. After this operation, all the edges in the path from 2 to 6 have a weight of 2. Hence, the answer is 1.
In the fourth query, we change the weights of edges [0,1], [1,2] and [2,3] to 2. After these operations, all the edges in the path from 0 to 6 have a weight of 2. Hence, the answer is 3.
For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from a<sub>i</sub> to b<sub>i</sub>.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/08/11/graph-9-1.png" style="width: 472px; height: 370px;" />
<pre>
<strong>Input:</strong> n = 8, edges = [[1,2,6],[1,3,4],[2,4,6],[2,5,3],[3,6,6],[3,0,8],[7,0,2]], queries = [[4,6],[0,4],[6,5],[7,4]]
<strong>Output:</strong> [1,2,2,3]
<strong>Explanation:</strong> In the first query, we change the weight of edge [1,3] to 6. After this operation, all the edges in the path from 4 to 6 have a weight of 6. Hence, the answer is 1.
In the second query, we change the weight of edges [0,3] and [3,1] to 6. After these operations, all the edges in the path from 0 to 4 have a weight of 6. Hence, the answer is 2.
In the third query, we change the weight of edges [1,3] and [5,2] to 6. After these operations, all the edges in the path from 6 to 5 have a weight of 6. Hence, the answer is 2.
In the fourth query, we change the weights of edges [0,7], [0,3] and [1,3] to 6. After these operations, all the edges in the path from 7 to 4 have a weight of 6. Hence, the answer is 3.
For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from a<sub>i</sub> to b<sub>i</sub>.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; n</code></li>
	<li><code>1 &lt;= w<sub>i</sub> &lt;= 26</code></li>
	<li>The input is generated such that <code>edges</code> represents a valid tree.</li>
	<li><code>1 &lt;= queries.length == m &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
</ul>

class Solution:
    def lcaHelper(self, tree, parent, dep, up, curNode, preNode):
        parent[curNode][0] = preNode
        dep[curNode] = dep[preNode] + 1

        for i in range(1, up + 1):
            parent[curNode][i] = parent[parent[curNode][i - 1]][i - 1]

        for nxtNode in tree[curNode]:
            if nxtNode == preNode:
                continue
            self.lcaHelper(tree, parent, dep, up, nxtNode, curNode)


    def lca(self, tree, parent, dep, up, node1, node2):
        if dep[node1] > dep[node2]:
            node1, node2 = node2, node1
        
        curDepDiff = dep[node2] - dep[node1]
        for i in range(up + 1):
            if curDepDiff & (1 << i):
                node2 = parent[node2][i]
        
        if node1 == node2:
            return node1

        for i in range(up, -1, -1):
            if parent[node1][i] != parent[node2][i]:
                node1 = parent[node1][i]
                node2 = parent[node2][i]
        return parent[node1][0]


    def minOperationsQueries(self, n: int, edges: List[List[int]], queries: List[List[int]]) -> List[int]:
        tree = collections.defaultdict(dict)
        for edge in edges:
            u, v, w = edge
            u += 1; v += 1
            tree[u][v] = tree[v][u] = w
        
        MAXV = 26
        MAXD = 20
        que = collections.deque([1])
        seen = {1}
        preSum = [[0] * (MAXV + 1) for _ in range(n + 1)]
        while que:
            curNode = que.popleft()
            for nxtNode in tree[curNode]:
                if nxtNode not in seen:
                    que.append(nxtNode)
                    seen.add(nxtNode)
                    value = tree[curNode][nxtNode]
                    for i in range(1, MAXV + 1):
                        preSum[nxtNode][i] = preSum[curNode][i]
                    preSum[nxtNode][value] += 1

        parent, dep = [[0] * (MAXD + 1) for _ in range(n + 1)], [0] * (n + 1)
        self.lcaHelper(tree, parent, dep, MAXD, 1, 0)

        m = len(queries)
        res = [n] * m
        for i in range(m):
            a, b = queries[i]
            a += 1; b += 1
            lca = self.lca(tree, parent, dep, MAXD, a, b)

            total = 0
            for j in range(1, MAXV + 1):
                total += preSum[a][j] + preSum[b][j] - 2 * preSum[lca][j]
            for j in range(1, MAXV + 1):
                curSum = preSum[a][j] + preSum[b][j] - 2 * preSum[lca][j]
                res[i] = min(res[i], total - curSum)
        return res
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