<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>

#include <unordered_map>
#include <unordered_set>
#include <vector>
#include <queue>

using namespace std;

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
private:
    // Helper function to populate parent map using BFS
    void markParents(TreeNode* root, unordered_map<TreeNode*, TreeNode*>& parent) {
        queue<TreeNode*> q;
        q.push(root);

        while (!q.empty()) {
            TreeNode* curr = q.front();
            q.pop();

            if (curr->left) {
                parent[curr->left] = curr; // Map child to parent
                q.push(curr->left);
            }

            if (curr->right) {
                parent[curr->right] = curr; // Map child to parent
                q.push(curr->right);
            }
        }
    }

public:
    int countPairs(TreeNode* root, int distance) {
        unordered_map<TreeNode*, TreeNode*> parent; // Map to store parent-child relationships
        unordered_set<TreeNode*> leaves; // Set to store all leaf nodes

        // Step 1: Build the parent map and identify all leaf nodes
        markParents(root, parent);

        // Traverse the tree to collect all leaf nodes
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            TreeNode* curr = q.front();
            q.pop();

            if (!curr->left && !curr->right) {
                leaves.insert(curr);
            }

            if (curr->left) q.push(curr->left);
            if (curr->right) q.push(curr->right);
        }

        int count = 0; // Count of good node pairs

        // Step 2: BFS from each leaf node to find other leaves within distance
        for (auto& leaf : leaves) {

            queue<TreeNode*> bfsQueue;
            unordered_set<TreeNode*> visited;
            bfsQueue.push(leaf);
            visited.insert(leaf);

            int level = 0; // Level indicator for BFS
            while (!bfsQueue.empty()) {
                int size = bfsQueue.size();
                if (level++ == distance) break; // Stop if we reach distance k

                for (int i = 0; i < size; ++i) {
                    TreeNode* curr = bfsQueue.front();
                    bfsQueue.pop();

                    // Explore neighbors (left child, right child, and parent)
                    if (curr->left && !visited.count(curr->left)) {
                        visited.insert(curr->left);
                        bfsQueue.push(curr->left);
                    }
                    if (curr->right && !visited.count(curr->right)) {
                        visited.insert(curr->right);
                        bfsQueue.push(curr->right);
                    }
                    if (parent[curr] && !visited.count(parent[curr])) {
                        visited.insert(parent[curr]);
                        bfsQueue.push(parent[curr]);
                    }
                }
            }

            // Count the number of other leaves within the distance
            for (const auto& node : visited) {
                if (leaves.count(node) && node != leaf) { //is it leaf? and node should not be itself.
                    ++count;
                }
            }
        }

        // Each pair is counted twice (once for each node as the start of BFS), so divide by 2
        return count / 2;
    }
};

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