一、题目

在这里插入图片描述

二、解法

思路分析：这道题关键在于分析插入值的位置，不论插入的值是什么（插入值和原有树中的键值都不相等），最终都是在空节点的位置插入，那么我们就可以确定递归的终止条件为空节点。因此只要和中间节点比较键值，确定递归是左子树还是右子树，递归完成后返回根节点。
程序如下：

class Solution {
public: 
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == NULL) {
            TreeNode* cur = new TreeNode(val);
            return cur;
        }      
        if (root->val > val) root->left = insertIntoBST(root->left, val);
        if (root->val < val) root->right = insertIntoBST(root->right, val);
        return root;
    }
};

三、完整代码

# include <iostream>
# include <vector>
# include <string>
# include <queue>
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: 
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == NULL) {
            TreeNode* cur = new TreeNode(val);
            return cur;
        }      
        if (root->val > val) root->left = insertIntoBST(root->left, val);
        if (root->val < val) root->right = insertIntoBST(root->right, val);
        return root;
    }
};

// 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()
{
    // 构建二叉树
    vector<string> t = { "4", "2", "1", "NULL", "NULL", "3", "NULL", "NULL", "7", "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:");

    // 插入目标值
    int val = 5;
    Solution s;
    TreeNode* result = s.insertIntoBST(root, val);
    vector<vector<int>> tree1 = levelOrder(result);
    my_print2<vector<vector<int>>, vector<int>>(tree1, "目标树:");

    system("pause");
    return 0;
}