给一个二叉树和目标值,判断是否存在从根节点到叶子节点的乘积等于目标值的路径;
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
class TreeNode {
public:
int val;
TreeNode *left;
TreeNode *right;
TreeNode() :val(0),left(nullptr),right(nullptr){}
TreeNode(int value) :val(value), left(nullptr), right(nullptr) {}
};
void dfs(TreeNode *a, int num,bool &flag,const int target) {
if (a == nullptr) {
return;
}
num = a->val*num;
if (a->left == nullptr&&a->right == nullptr&&num == target) {
flag = true;
}
if (a->left) {
dfs(a->left, num,flag, target);
}
if (a->right) {
dfs(a->right, num,flag, target);
}
}
int main() {
TreeNode *a = new TreeNode(2);
TreeNode *b = new TreeNode(1);
TreeNode *c = new TreeNode(3);
TreeNode *d = new TreeNode(4);
TreeNode *e = new TreeNode(6);
TreeNode *f = new TreeNode(7);
a->left = b;
a->right = c;
b->left = d;
c->left = e;
c->right = f;
bool flag = false;
int target;
cout << "请输入目标值:";
cin >> target;
int len = 1;
dfs(a,len,flag,target);
cout<< boolalpha <<flag<< endl;;
system("pause");
}