题目描述
输入一颗二叉树的根节点和一个整数,按字典序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。
解:该题目如果不考虑输入整数问题就是一个打印二叉树路径的问题,打印二叉树无非就是对树进行遍历;加入和某一整数相等的路径筛选只要做少许改动即可,具体程序实现如下,代码根据自己思想写的,可能有些冗余。
/*
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> > FindPath(TreeNode* root,int expectNumber) {
vector<vector<int>> result;
if(root == NULL)
return result;
vector<int> path;
path.push_back(root -> val);
int sum = root -> val;
DFS(root,expectNumber,result,path,sum);
return result;
}
void DFS(TreeNode *root,int expectNumber,vector<vector<int>> &result, vector<int> path,int sum)
{
if(root -> left == NULL && root -> right == NULL && sum == expectNumber)
{
result.push_back(path);
return;
}
if(root -> left == NULL && root -> right == NULL && sum != expectNumber)
return;
int sum1 = sum;
vector<int> path1 = path;
if(root -> left != NULL)
{
sum += root -> left -> val;
path.push_back(root -> left -> val);
DFS(root -> left,expectNumber,result,path,sum);
}
if(root -> right != NULL)
{
sum1 += root -> right -> val;
path1.push_back(root -> right -> val);
DFS(root -> right,expectNumber,result,path1,sum1);
}
}
};