<p>You are given the <code>root</code> of a binary tree and an integer <code>distance</code>. A pair of two different <strong>leaf</strong> nodes of a binary tree is said to be good if the length of <strong>the shortest path</strong> between them is less than or equal to <code>distance</code>.</p>

<p>Return <em>the number of good leaf node pairs</em> in the tree.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/09/e1.jpg" style="width: 250px; height: 250px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4], distance = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/09/e2.jpg" style="width: 250px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6,7], distance = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of ther shortest path between them is 4.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only good pair is [2,5].
</pre>

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

<ul>
	<li>The number of nodes in the <code>tree</code> is in the range <code>[1, 2<sup>10</sup>].</code></li>
	<li><code>1 &lt;= Node.val &lt;= 100</code></li>
	<li><code>1 &lt;= distance &lt;= 10</code></li>
</ul>

class Solution {
public:
    unordered_map<TreeNode*, TreeNode*> parent;
    unordered_map<TreeNode*, int> depth;
    vector<TreeNode *> leafs;

    void preprocess(TreeNode* root, TreeNode* par = nullptr, int d = 0) {
        if (root == nullptr) return;
        parent[root] = par;
        depth[root] = d;
        if (root->left == nullptr && root->right == nullptr) {
            leafs.push_back(root);
        }
        preprocess(root->left, root, d + 1);
        preprocess(root->right, root, d + 1);
    }

    TreeNode* lca(TreeNode* node1, TreeNode* node2) {
        if (depth[node1] < depth[node2]) swap(node1, node2);
        while (depth[node1] > depth[node2]) {
            node1 = parent[node1];
        }
        while (node1 != node2) {
            node1 = parent[node1];
            node2 = parent[node2];
        }
        return node1;
    }

    int dist(TreeNode* ances, TreeNode* leaf) {
        int d = 0;
        while (leaf != ances) {
            leaf = parent[leaf];
            d++;
        }
        return d;
    }

    int countPairs(TreeNode* root, int distance) {
        int goodpairs = 0;
        preprocess(root);

        int n = leafs.size();
        vector<vector<int>> distMatrix(n, vector<int>(n, 0));
        
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                TreeNode* ances = lca(leafs[i], leafs[j]);
                int dist1 = dist(ances, leafs[i]);
                int dist2 = dist(ances, leafs[j]);
                distMatrix[i][j] = distMatrix[j][i] = dist1 + dist2;
            }
        }
        
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (distMatrix[i][j] <= distance) {
                    goodpairs++;
                }
            }
        }
        
        return goodpairs;
    }
};

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