1530. Number of Good Leaf Nodes Pairs

You are given the root of a binary tree and an integer distance. A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path between them is less than or equal to distance.

Return the number of good leaf node pairs in the tree.

Example 1:

Input: root = [1,2,3,null,4], distance = 3
Output: 1
Explanation: 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.

Example 2:

Input: root = [1,2,3,4,5,6,7], distance = 3
Output: 2
Explanation: 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.

Example 3:

Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
Output: 1
Explanation: The only good pair is [2,5].

题目：给定一个二叉树，找出good pair的数量，good pair是距离路径长度小于给定值distance的叶子对。

思路：叶子不可能直接与叶子相通，必然是通过一个共同的父结点或祖辈节点，需要记录以每个节点作为根节点的子树下有几个叶子节点，且叶子节点离当前子根节点的距离。可递归为计算其左节点下的叶子节点和其右节点下的叶子节点，计算其所有左叶子节点和所有右叶子节点的距离（为其左叶子节点距离当前节点的距离+右叶子节点距离当前节点的距离），如距离小于给定distance，则更新结果。最后将其所有左叶子节点和所有右叶子节点整合，为当前节点的所有子叶子节点。用hashmap记录。

注：hasmap的key记录当前子节点所有叶子节点的距离，value记录某一距离的叶子节点的个数

代码：

/**
 * 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:
    unordered_map helper(TreeNode* node, int dis, int& res){
        if(!node) return {};
        unordered_map left = helper(node->left, dis, res);
        unordered_map right = helper(node->right, dis, res);
        unordered_map out;
        if(left.empty() && right.empty()){
            out[0] = 1;
        } else {
            for(auto l : left){
                out[l.first + 1] += l.second;
                for(auto r : right){
                    if (l.first + r.first + 2 <= dis){
                        res += l.second * r.second;
                    }
                }
            }
            for(auto r : right){
                out[r.first+1] += r.second;
            }
        }
        return out;
    }
    int countPairs(TreeNode* root, int distance) {
        int res = 0;
        helper(root, distance, res);
        return res;
    }
};