二叉树的遍历是最基础的算法，也是比较简单的算法，作为一名大学生这是我们必须掌握的基础知识。因为遍历的思路很简单，所以今天这篇是纯代码篇，我们直接在代码中学习。
目录
一、二叉树前序遍历
题目链接：力扣
前序遍历的遍历顺序：根 左 右
所以这道题就很简单了，我们可以写出递归版本和迭代版本的代码。
递归版：
/**
* Definition for a binary tree node.
* 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:
vector<int> res;
vector<int> preorderTraversal(TreeNode* root) {
dfs(root);
return res;
}
void dfs(TreeNode* root){
if(root == nullptr) return;
res.push_back(root->val);
dfs(root->left);
dfs(root->right);
}
};迭代版：
其实，递归版本和迭代版本的思路是一样的，我们只是需要用一个栈来模拟递归的过程即可。
/**
* Definition for a binary tree node.
* 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:
vector<int> preorderTraversal(TreeNode* root) {
stack<TreeNode*> st;
vector<int> res;
TreeNode* cur = root;
while(cur != nullptr || !st.empty()){
if(cur != nullptr){
st.push(cur);
res.push_back(cur->val);
cur = cur->left;
}
else{
cur = st.top();
st.pop();
cur = cur->right;
}
}
return res;
}
};二、二叉树的中序遍历
中序遍历遍历顺序：左 根 右
递归版：
/**
* Definition for a binary tree node.
* 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:
vector<int> res;
vector<int> inorderTraversal(TreeNode* root) {
dfs(root);
return res;
}
void dfs(TreeNode* root){
if(root == nullptr) return;
dfs(root->left);
res.push_back(root->val);
dfs(root->right);
}
};迭代版：
/**
* Definition for a binary tree node.
* 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:
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode*> st;
vector<int> res;
TreeNode* cur = root;
while(cur != nullptr || !st.empty()){
if(cur!=nullptr){
st.push(cur);
cur = cur->left;
}
else{
cur = st.top();
st.pop();
res.push_back(cur->val);
cur = cur->right;
}
}
return res;
}
};三、二叉树的后序遍历
题目链接：力扣
递归版：
/**
* Definition for a binary tree node.
* 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:
vector<int> res;
vector<int> postorderTraversal(TreeNode* root) {
dfs(root);
return res;
}
void dfs(TreeNode* root){
if(root == nullptr) return;
dfs(root->left);
dfs(root->right);
res.push_back(root->val);
}
};迭代版：
/**
* Definition for a binary tree node.
* 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:
vector<int> postorderTraversal(TreeNode* root) {
stack<TreeNode*> st;
vector<int> res;
TreeNode* cur = root;
while(cur != nullptr || !st.empty()){
if(cur!=nullptr){
st.push(cur);
res.push_back(cur->val);
cur = cur->right;
}
else{
cur = st.top();
st.pop();
cur = cur->left;
}
}
reverse(res.begin(),res.end());
return res;
}
};四、二叉树的层序遍历
层序遍历一个二叉树。就是从左到右一层一层的去遍历二叉树。这种遍历的方式和我们之前讲过的都不太一样。
这时候我们需要借用一个辅助数据结构即队列来实现，队列先进先出，符合一层一层遍历的逻辑，而是用栈先进后出适合模拟深度优先遍历也就是递归的逻辑。
而这种层序遍历方式就是图论中的广度优先遍历，只不过我们应用在二叉树上。
/**
* Definition for a binary tree node.
* 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:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
queue<TreeNode*> que;
if(root!=nullptr) que.push(root);
while(!que.empty()){
int size = que.size();
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);
}
res.push_back(vec);
}
return res;
}
};