<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>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
map<TreeNode*, int >m;
vector<int> v[2050];
int node,ans;
    int countPairs(TreeNode* root, int distance) {
        m.clear();
        for(int i=0;i<=2049;i++) {
            v[i].clear();
            for(int j=0;j<=distance;j++)
            v[i].push_back(0);
        }
        node=1;
        ans=0;
        dfs(root,distance);
        calculate(root,distance);
        return ans;
    }

    void dfs(TreeNode* root, int distance) {
        if(root==NULL)
        return;
        if(m.find(root)==m.end()) {
            m[root]=node++;
        }
        dfs(root->left,distance);
        dfs(root->right,distance);
        if(root->left==NULL && root->right==NULL) {
            v[m[root]][0]=1;
        }
        if(root->left!=NULL) {
            for(int i=1;i<=distance;i++) {
                v[m[root]][i]+=v[m[root->left]][i-1]+v[m[root->left]+1025][i-1];
            }
        }
        if(root->right!=NULL) {
            for(int i=1;i<=distance;i++) {
                v[m[root]+1025][i]+=v[m[root->right]][i-1]+v[m[root->right]+1025][i-1];
            }
        }
    }

    void calculate(TreeNode* root, int distance) {
        if(root==NULL)
        return;
        int val=m[root];
        int sum=0;
        for(int i=0;i<=distance;i++)
        sum+=v[val+1025][i];
        for(int i=0;i<=distance;i++) {
            ans+=v[val][i]*sum;
            sum-=v[val+1025][distance-i];
        }
        calculate(root->left,distance);
        calculate(root->right,distance);
    }
};
76493