<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:
    // 对于 dfs(root,distance)，同时返回：
    // 1）每个叶子节点与 root 之间的距离
    // 2) 以 root 为根节点的子树中好叶子节点对的数量
    pair<vector<int>, int> dfs(TreeNode* root, int distance) {
        // 超过 distance + 1 的叶子节点不再统计
        vector<int> depths(distance + 1, 0);
        bool isLeaf = (!root->left && !root->right);
        if (isLeaf) { 
            depths[0] = 1; // itself
            return make_pair(depths, 0);
        }

        vector<int> leftDepths(distance + 1, 0), rightDepths(distance + 1, 0);
        int leftCount = 0, rightCount = 0;
        if (root->left) {
            tie(leftDepths, leftCount) = dfs(root->left, distance);
        }
        if (root->right) {
            tie(rightDepths, rightCount) = dfs(root->right, distance);
        }

        for (int i = 0; i < distance; i++) {
            // depth(father)=depth(child)+1
            depths[i + 1] += leftDepths[i];
            depths[i + 1] += rightDepths[i];
        }

        // whoa! great idea here, simple combination
        int cnt = 0;
        for (int i = 0; i <= distance; i++) {
            for (int j = 0; j + i + 2 <= distance; j++) {
                cnt += (leftDepths[i] * rightDepths[j]);
            }
        }
        return make_pair(depths, cnt + leftCount + rightCount);
    }

    int countPairs(TreeNode* root, int distance) {
        auto [depths, ret] = dfs(root, distance);
        return ret;
    }
};
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