DFS遍历整棵树，记录每个节点到所有左右子树叶子节点的距离。
dist[i][j][k] 表示第i个节点左子树(j = 0)或右子树(j = 1)距离为k的叶子节点个数
最后用双指针计算每个节点满足条件的叶子对。
算法复杂度O(n*distance)

class Solution {
public:

    int dist[1100][2][15];
    int cnt = 0;

    void dfs(TreeNode* node, int father, bool isRight){
        if (!node)  return;
        cnt++;
        int idx = cnt;
        dfs(node->left, idx, 0);
        dfs(node->right, idx, 1);

        //judge leaf node
        if (!node->left && !node->right)    dist[idx][0][0]++;
        for (int i = 0; i<=10; i++)
            dist[father][isRight][i+1] = dist[idx][0][i] + dist[idx][1][i];
    }

    int countPairs(TreeNode* root, int distance) {
        memset(dist, 0, sizeof dist);
        dfs(root, 0, 0);

        int ans = 0;
        for (int i = 1; i<=cnt; i++){
            int sum = 0;
            for (int j = distance - 1, k = 1; j >=1; j--, k++){
                sum += dist[i][1][k];
                ans += dist[i][0][j] * sum;
            }
        }
        return ans;
    }
};