所有的LeetCode题解索引，可以看这篇文章——【算法和数据结构】LeetCode题解。

一、题目

二、解法

  思路分析：这道题的思路很简单，本质上就是遍历每一个节点，然后交换左右节点。我们可以用前中后遍历或者是层次遍历法来做，参考这两篇文章，【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）和【算法和数据结构】102、LeetCode二叉树的层序遍历。交换函数可以用库函数也可以自己写一个my_sawp()函数。
  my_sawp()函数如下：

void my_swap(TreeNode* &root) {TreeNode* node = root->left;root->left = root->right;root->right = node;
}

  迭代法前序遍历程序如下：

class Solution1 {
public:// 迭代法,前序遍历TreeNode* invertTree(TreeNode* root) {stack<TreeNode*> st;if (root != NULL) st.push(root);while (!st.empty()) {TreeNode* node = st.top();              // 中st.pop();swap(node->left, node->right);if (node->right) st.push(node->right);  // 右if (node->left) st.push(node->left);    // 左}return root;}
};

  前序遍历，统一代码风格迭代写法：

class Solution2 {
public:// 前序遍历，统一代码风格迭代写法TreeNode* invertTree(TreeNode* root) {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();swap(node->left, node->right);}}return root;}
};

  递归法程序如下：

class Solution3 {
public:// 递归法TreeNode* invertTree(TreeNode* root) {if (root == NULL) return root;swap(root->left, root->right);invertTree(root->left);invertTree(root->right);return root;}
};

  层序遍历法程序如下：

class Solution4 {
public:// 层序遍历法TreeNode* invertTree(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for (int i = 0; i < size; ++i) {TreeNode* node = que.front();swap(node->left, node->right);que.pop();if (node->left) que.push(node->left);if (node->right) que.push(node->right);}}return root;}
};

三、完整代码

# include <iostream>
# include <vector>
# include <queue>
# include <string>
# 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) {}
};void my_swap(TreeNode*& root) {TreeNode* node = root->left;root->left = root->right;root->right = node;
}class Solution1 {
public:// 迭代法,前序遍历TreeNode* invertTree(TreeNode* root) {stack<TreeNode*> st;if (root != NULL) st.push(root);while (!st.empty()) {TreeNode* node = st.top();              // 中st.pop();swap(node->left, node->right);if (node->right) st.push(node->right);  // 右if (node->left) st.push(node->left);    // 左}return root;}
};class Solution2 {
public:// 前序遍历，统一代码风格迭代写法TreeNode* invertTree(TreeNode* root) {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();swap(node->left, node->right);}}return root;}
};class Solution3 {
public:// 递归法TreeNode* invertTree(TreeNode* root) {if (root == NULL) return root;swap(root->left, root->right);invertTree(root->left);invertTree(root->right);return root;}
};class Solution4 {
public:// 层序遍历法TreeNode* invertTree(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for (int i = 0; i < size; ++i) {TreeNode* node = que.front();swap(node->left, node->right);que.pop();if (node->left) que.push(node->left);if (node->right) que.push(node->right);}}return root;}
};void my_print2(vector<vector<int>>& v, string str) {cout << str << endl;for (vector<vector<int>>::iterator vit = v.begin(); vit < v.end(); ++vit) {for (vector<int>::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {cout << *it << ' ';}cout << endl;}
}// 前序遍历递归法创建二叉树，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[0] == "NULL" || !t.size()) return;    // 退出条件else {node = new TreeNode(stoi(t[0].c_str()));    // 中t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->left);              // 左t.assign(t.begin() + 1, t.end());Tree_Generator(t, node->right);             // 右}
}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> t = { "4", "2", "1", "NULL", "NULL", "3", "NULL", "NULL", "7", "6", "NULL", "NULL", "9", "NULL", "NULL" };   // 前序遍历TreeNode* root = new TreeNode();Tree_Generator(t, root);vector<vector<int>> tree = levelOrder(root);my_print2(tree, "目标树:");Solution2 s1;root = s1.invertTree(root);tree = levelOrder(root);my_print2(tree, "翻转后：");system("pause");return 0;
}

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