二叉树中和为某一值的路径

题目描述

输入一颗二叉树和一个整数，打印出二叉树中结点值的和为输入整数的所有路径。路径定义为从树的根结点开始往下一直到叶结点所经过的结点形成一条路径。算法：需要从根节点值开始，所有dfs中只有前序遍历满足条件使用path数组记录路径，每次到遍历完一个节点的最后应该pop_back()，并返回上一层。代码

#include <iostream>
#include <vector>
using namespace std;

struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
class Solution {
public:
vector<vector<int> > res;

vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {

if(root==NULL)
return res;
vector<int> path;
preFind(root,expectNumber,path);
return res;

}

void preFind(TreeNode* root,int expectNumber,vector<int> path){
path.push_back(root->val);
if(root->left!=NULL){
preFind(root->left,expectNumber,path);
}
if(root->right!=NULL){
preFind(root->right,expectNumber,path);
}
int sum = 0;
for (int i = 0; i < path.size(); ++i) {
sum += path[i];
}
if(sum == expectNumber && root->right==NULL && root->left==NULL){
res.push_back(path);
}
path.pop_back();
sum = 0;
}
};
int main() {
TreeNode *root= new TreeNode(10);
TreeNode *r1l = new TreeNode(5);
TreeNode *r1r = new TreeNode(12);
TreeNode *r2l = new TreeNode(4);
TreeNode *r2r = new TreeNode(7);
root->left = r1l;
root->right = r1r;
r1l->left = r2l;
r1l->right = r2r;
vector<vector<int>>res;
Solution s;
res = s.FindPath(root,22);

cout << "Hello, World!" << endl;
return 0;
}