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：文章来源：https://www.toymoban.com/news/detail-603625.html
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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