一、题目

在这里插入图片描述

二、解法

思路分析：本题通过计算根节点到叶子节点路径上节点的值之和，然后再对比目标值。利用文章【算法和数据结构】257、LeetCode二叉树的所有路径中的递归算法。这里要注意，默认路径之和是不等于目标值，一旦递归当中出现了等于的情况就直接返回，不必继续算后面的和。因此程序当中将结果result作为引用输入参数，有true出现就直接退出了。
程序如下：

class Solution {
public:          void traversal(TreeNode* root, int sumOfPath, const int targetSum, bool &result) {// 1.输入参数和返回值 sumOfPath += root->val;// 2.终止条件：遇到叶子节点if (!root->left && !root->right) {if (sumOfPath == targetSum) result = true;}// 3.单层递归逻辑：递归+回溯if (root->left && !result)  traversal(root->left, sumOfPath, targetSum, result);    // 左                         if (root->right && !result) traversal(root->right, sumOfPath, targetSum, result);  // 右}bool hasPathSum(TreeNode* root, int targetSum) {bool result = false;if(root) traversal(root, 0, targetSum, result);return result;}
};

复杂度分析：

时间复杂度： $O (n)$ 。
空间复杂度： $O (n)$ 。

三、完整代码

# include <iostream>
# include <vector>
# include <queue>
# include <string>
# include <algorithm>
# include <stack>
using namespace std;// 树节点定义
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 {
public:          void traversal(TreeNode* root, int sumOfPath, const int targetSum, bool &result) {// 1.输入参数和返回值 sumOfPath += root->val;// 2.终止条件：遇到叶子节点if (!root->left && !root->right) {if (sumOfPath == targetSum) result = true;}// 3.单层递归逻辑：递归+回溯if (root->left && !result)  traversal(root->left, sumOfPath, targetSum, result);    // 左                         if (root->right && !result) traversal(root->right, sumOfPath, targetSum, result);  // 右}bool hasPathSum(TreeNode* root, int targetSum) {bool result = false;if(root) traversal(root, 0, targetSum, result);return result;}
};template<typename T>
void my_print(T& v, const string msg)
{cout << msg << endl;for (class T::iterator it = v.begin(); it != v.end(); it++) {cout << *it << ' ';}cout << endl;
}template<class T1, class T2>
void my_print2(T1& v, const string str) {cout << str << endl;for (class T1::iterator vit = v.begin(); vit < v.end(); ++vit) {for (class T2::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {cout << *it << ' ';}cout << endl;}
}// 前序遍历迭代法创建二叉树，每次迭代将容器首元素弹出（弹出代码还可以再优化）
void Tree_Generator(vector<string>& t, TreeNode*& node) {if (!t.size() || t[0] == "NULL") return;    // 退出条件else {node = new TreeNode(stoi(t[0].c_str()));    // 中if (t.size()) {t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->left);              // 左}if (t.size()) {t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->right);             // 右}}
}// 层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {queue<TreeNode*> que;if (root != NULL) que.push(root);vector<vector<int>> result;while (!que.empty()) {int size = que.size();  // size必须固定, que.size()是不断变化的vector<int> vec;for (int i = 0; i < size; ++i) {TreeNode* node = que.front();que.pop();vec.push_back(node->val);if (node->left) que.push(node->left);if (node->right) que.push(node->right);}result.push_back(vec);}return result;
}// 二叉树所有路径
class Solution2 {
public:// 前序遍历递归法:精简版本      void traversal(TreeNode* root, string path, vector<string>& result) { // 1.输入参数和返回值        path += to_string(root->val);      // 中间节点先加入pathif (!root->left && !root->right) {  // 2.终止条件：遇到叶子节点result.push_back(path);return;}// 3.单层递归逻辑：递归+回溯if (root->left) traversal(root->left, path + "->", result);     // 左if (root->right) traversal(root->right, path + "->", result);   // 右}vector<string> binaryTreePaths(TreeNode* root) {vector<string> result;if (!root) return result;traversal(root, "", result);return result;}
};int main()
{vector<string> t = { "5", "4", "11", "7", "NULL", "NULL", "2", "NULL", "NULL", "NULL", "8", "13", "NULL", "NULL", "4", "NULL", "1", "NULL", "NULL"};   // 前序遍历my_print(t, "目标树");TreeNode* root = new TreeNode();Tree_Generator(t, root);vector<vector<int>> tree = levelOrder(root);my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");Solution2 s2;vector<string> path = s2.binaryTreePaths(root);my_print(path, "所有路径为：");Solution s;int targetSum = 22;bool result = s.hasPathSum(root, targetSum);cout << "路径总和是否满足目标值：  " << result << endl;system("pause");return 0;
}