给定一个二叉树的根节点 root ,和一个整数 targetSum ,求该二叉树里节点值之和等于 targetSum 的 路径 的数目。
路径 不需要从根节点开始,也不需要在叶子节点结束,但是路径方向必须是向下的(只能从父节点到子节点)。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/path-sum-iii
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输入:root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出:3
解释:和等于 8 的路径有 3 条,如图所示。
示例 2:
输入:root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出:3
提示:
二叉树的节点个数的范围是 [0,1000]
-109 <= Node.val <= 109
-1000 <= targetSum <= 1000
#include
#include
#include
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) {}
};
typedef struct TreeNode Node;
class Solution {
public:
void traverseAndCount(TreeNode* root,long& targetSum,int& count)
{
if (!root)
return;
targetSum -= root->val;
if (targetSum == 0)
count+=1;
if (root->left)
{
traverseAndCount(root->left,targetSum,count);
targetSum += root->left->val;
}
if (root->right)
{
traverseAndCount(root->right,targetSum,count);
targetSum += root->right->val;
}
}
void getAllNode(std::stack<TreeNode*>& nodeStack,TreeNode* root)
{
if (root)
{
nodeStack.push(root);
getAllNode(nodeStack,root->left);
getAllNode(nodeStack,root->right);
}
}
int pathSum(TreeNode* root, int targetSum) {
std::stack<TreeNode*> nodeStack;
getAllNode(nodeStack,root);
int count = 0;
while (!nodeStack.empty())
{
auto node = nodeStack.top();
long increVal = targetSum;
traverseAndCount(node,increVal,count);
nodeStack.pop();
}
return count;
}
};
int main()
{
Solution slu;
Node root(3);
Node node1(2);
Node node2(3);
Node node3(3);
Node node4(1);
root.left = &node1;
root.right= &node2;
node1.right = &node3;
node2.right = &node4;
std::cout << slu.pathSum(&root,5) << std::endl;
}