<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>

// N=64 version dealing only for tree height<64
class Solution {
public:
    static const int N = 64;
    using info = pair<uint16_t, bitset<N>>; // (level, code)

    static void dfs(TreeNode* root, uint16_t level, bitset<N>& code, vector<info>& leaf) {
        if (!root) return;
        if (!root->left && !root->right) {
            leaf.emplace_back(level, code); // leaf node with info (level, code)
        }
        if (root->left) {
            code[level] = 0;
            dfs(root->left, level+1, code, leaf);
        }
        if (root->right) {
            code[level] = 1;
            dfs(root->right, level+1, code, leaf);
            code[level] = 0; // backtrack
        }
    }

    static int countPairs(TreeNode* root, int distance) {
        vector<info> leaf;
        leaf.reserve(N);
        bitset<N> code=1;// root is encoded as 1
        dfs(root, 1, code, leaf);
        int sz = leaf.size(), cnt = 0;
        for (int i=0; i<sz-1; i++) {
            auto& [level0, code0] = leaf[i];
            for (int j = i+1; j < sz; j++) {
                auto& [level1, code1] = leaf[j];
                // Find the common prefix length (LCA)
                int len=0, minLevel=min(level0, level1);
                if (level0+level1-2*minLevel>distance) continue;
                unsigned long long diff=(code0^code1).to_ullong();
                len=min(minLevel, countr_zero(diff));
                // Calculate the distance using the levels and common length
                int d=(level0+level1-2*len);
                if (d <= distance) cnt++;
            }
        }
        return cnt;
    }
};






auto init = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 'c';
}();
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