一、题目

在这里插入图片描述

二、解法

思路分析：采用递归的方式遍历二叉树，【算法与数据结构】144、94、145LeetCode二叉树的前中后遍历（递归法、迭代法），递归法程序可以参考这篇文章。递归重要的是三步骤：输入参数和返回值；终止条件；单层递归逻辑。
程序如下：

class Solution {
public://1、 输入参数root1 root2TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {// 2、终止条件if (!root1) return root2;if (!root2) return root1;// 3、单层递归逻辑root1->val += root2->val;root1->left = mergeTrees(root1->left, root2->left);root1->right = mergeTrees(root1->right, root2->right);// 1、返回值 root1return root1;}
};

三、完整代码

# include <iostream>
# include <vector>
# include <string>
# include <queue>
# 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://1、 输入参数root1 root2TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {// 2、终止条件if (!root1) return root2;if (!root2) return root1;// 3、单层递归逻辑root1->val += root2->val;root1->left = mergeTrees(root1->left, root2->left);root1->right = mergeTrees(root1->right, root2->right);// 1、返回值 root1return root1;}
};// 前序遍历，统一代码风格迭代写法
class Solution8 {
public:vector<int> preorderTraversal(TreeNode* root) {vector<int> result;stack<TreeNode*> st;if (root != NULL) st.push(root);while (!st.empty()) {TreeNode* node = st.top();if (node != NULL) {st.pop();if (node->right) st.push(node->right);  // 右if (node->left) st.push(node->left);    // 左st.push(node);                          // 中st.push(NULL);}else {st.pop();node = st.top();st.pop();result.push_back(node->val);}}return result;}
};// 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}}
}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;}
}// 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;
}int main()
{vector<string> t1 = { "1", "3", "5", "NULL", "NULL", "NULL", "2", "NULL", "NULL"};   // 前序遍历my_print(t1, "目标树");TreeNode* root1 = new TreeNode();Tree_Generator(t1, root1);vector<vector<int>> tree1 = levelOrder(root1);my_print2<vector<vector<int>>, vector<int>>(tree1, "目标树:");vector<string> t2 = { "2", "1", "NULL", "4", "NULL", "NULL", "3", "NULL", "7", "NULL", "NULL" };   // 前序遍历my_print(t2, "目标树");TreeNode* root2 = new TreeNode();Tree_Generator(t2, root2);vector<vector<int>> tree2 = levelOrder(root2);my_print2<vector<vector<int>>, vector<int>>(tree2, "目标树:");Solution s;TreeNode* root = s.mergeTrees(root1, root2);vector<vector<int>> tree = levelOrder(root);my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");system("pause");return 0;
}