路径总和Ⅱ

与路径总和类似，只不过需要额外用数组来记录节点。

/**
 * 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 {
private:
    vector<vector<int>> result;
    
public:
    void dfs(TreeNode* root, vector<int> vec, int sum, int targetSum){
        sum += root->val;
        vec.push_back(root->val);

        if(root->left == nullptr && root->right == nullptr && sum == targetSum){
                result.push_back(vec);
        }

        if(root->left){
            dfs(root->left, vec, sum, targetSum);
        }
        if(root->right){
            dfs(root->right, vec, sum, targetSum);
        }
    }
    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        int sum = 0;
        vector<int> vec;
        if(root == nullptr){
            return result;
        }
        dfs(root, vec, sum, targetSum);
        return result;
    }
};

同样可以像路径总和一样，对代码进行处理，减少传入的参数，减少拷贝操作。

class Solution {
private:
    vector<vector<int>> result;
    vector<int> vec;
    
public:
    void dfs(TreeNode* root, int targetSum){
        targetSum -= root->val;
        vec.push_back(root->val);

        if(root->left == nullptr && root->right == nullptr && targetSum == 0){
                result.push_back(vec);
        }

        if(root->left){
            dfs(root->left, targetSum);
        }
        if(root->right){
            dfs(root->right, targetSum);
        }
        vec.pop_back();
    }
    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        vector<int> vec;
        if(root == nullptr){
            return result;
        }
        dfs(root, targetSum);
        return result;
    }
};