二叉树的前序、中序、后序、层序遍历
作者：互联网
1. 前序遍历
链接：https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
1.1 递归
代码：
class Solution {
public:
vector<int> v;
vector<int> preorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
v.push_back(root->val);
preorderTraversal(root->left);
preorderTraversal(root->right);
return v;
}
};
1.2 迭代
代码：
class Solution {
public:
vector<int> v;
vector<int> preorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
stack<TreeNode*> st;
st.push(root);
while(!st.empty())
{
TreeNode* tmp = st.top();
st.pop();
v.push_back(tmp->val);
if(tmp->right != nullptr)
st.push(tmp->right);
if(tmp->left != nullptr)
st.push(tmp->left);
}
return v;
}
};
1.3 morris前序遍历
代码：
class Solution {
public:
vector<int> v;
vector<int> preorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
TreeNode* predecessor = nullptr;
while(root != nullptr)
{
if(root->left != nullptr)
{
predecessor = root->left;
while(predecessor->right != nullptr && predecessor->right != root)
{
predecessor = predecessor->right;
}
if(predecessor->right == nullptr)
{
v.push_back(root->val);
predecessor->right = root;
root = root->left;
}
else
{
root = root->right;
predecessor->right = nullptr;
}
}
else
{
v.push_back(root->val);
root = root->right;
}
}
return v;
}
};
2.中序遍历
链接：https://leetcode-cn.com/problems/binary-tree-inorder-traversal/
2.1 递归
代码：
class Solution {
public:
vector<int> v;
vector<int> inorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
inorderTraversal(root->left);
v.push_back(root->val);
inorderTraversal(root->right);
return v;
}
};
2.2 迭代
2.2.1 改变树结构
代码：
class Solution {
public:
vector<int> v;
vector<int> inorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
stack<TreeNode*> st;
st.push(root);
while(!st.empty())
{
TreeNode* tmp = st.top();
if(tmp->left != nullptr)
{
st.push(tmp->left);
tmp->left = nullptr;
continue;
}
v.push_back(tmp->val);
st.pop();
if(tmp->right != nullptr)
st.push(tmp->right);
}
return v;
}
};
2.2.2 不改变树结构
代码：
class Solution {
public:
vector<int> v;
vector<int> inorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
stack<TreeNode*> st;
while(!st.empty() || root != nullptr)
{
while(root != nullptr)
{
st.push(root);
root = root->left;
}
root = st.top();
st.pop();
v.push_back(root->val);
root = root->right;
}
return v;
}
};
2.2.3 morris中序遍历
代码：
class Solution {
public:
vector<int> v;
vector<int> inorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
TreeNode* predecessor = nullptr;
while(root != nullptr)
{
if(root->left != nullptr)
{
predecessor = root->left;
while(predecessor->right != nullptr && predecessor->right != root)
{
predecessor = predecessor->right;
}
if(predecessor->right == nullptr)
{
predecessor->right = root;
root = root->left;
}
else
{
v.push_back(root->val);
predecessor->right = nullptr;
root = root->right;
}
}
else
{
v.push_back(root->val);
root = root->right;
}
}
return v;
}
};
3.后序遍历
链接：https://leetcode-cn.com/problems/binary-tree-postorder-traversal/
3.1 递归
代码：
class Solution {
public:
vector<int> v;
vector<int> postorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
postorderTraversal(root->left);
postorderTraversal(root->right);
v.push_back(root->val);
return v;
}
};
3.2迭代
代码：
class Solution {
public:
vector<int> v;
vector<int> postorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
stack<TreeNode*> st;
TreeNode* prev = nullptr;
while(root != nullptr || !st.empty())
{
while(root != nullptr)
{
st.push(root);
root = root->left;
}
root = st.top();
if(root->right == nullptr || root->right == prev)
{
v.push_back(root->val);
st.pop();
prev = root;
root = nullptr;
}
else
root = root->right;
}
return v;
}
};
3.3 morris后序遍历
代码：
class Solution {
public:
vector<int> v;
vector<int> postorderTraversal(TreeNode* root) {
if(root == nullptr)
return {};
TreeNode* newnode = new TreeNode(-1);//新建临时节点
newnode->left = root;
TreeNode* p = newnode;
while(p != nullptr)
{
if(p->left != nullptr)
{
TreeNode* predecessor = p->left;
while(predecessor->right != nullptr && predecessor->right != p)
predecessor = predecessor->right;
if(predecessor->right == nullptr)
{
predecessor->right = p;
p = p->left;
}
else
{
predecessor->right = nullptr;
_postorderTraversal(p->left);
p = p->right;
}
}
else
p = p->right;
}
delete newnode;
newnode = nullptr;
return v;
}
private:
void _postorderTraversal(TreeNode* root)
{
int count = 0;
while(root != nullptr)
{
++count;
v.push_back(root->val);
root = root->right;
}
reverse(v.end() - count, v.end());
}
};
4.层序遍历
链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
4.1 递归
代码：
class Solution {
public:
vector<vector<int>> vv;
vector<vector<int>> levelOrder(TreeNode* root) {
if(root == nullptr)
return vv;
_levelOrder(root, 0);
return vv;
}
private:
void _levelOrder(TreeNode* root, int index)
{
if(root == nullptr)
return;
if(index >= vv.size())
vv.push_back(vector<int>());
vv[index].push_back(root->val);
_levelOrder(root->left, index + 1);
_levelOrder(root->right, index + 1);
}
};
4.2 迭代
代码：
class Solution {
public:
vector<vector<int>> vv;
vector<vector<int>> levelOrder(TreeNode* root) {
if(root == nullptr)
return vv;
queue<TreeNode*> _que;
_que.push(root);
int count = 0;
while(!_que.empty())
{
int sz = _que.size();
vv.push_back(vector<int>());
for(int i = 0; i < sz; ++i)
{
TreeNode* tmp = _que.front();
_que.pop();
vv[count].push_back(tmp->val);
if(tmp->left != nullptr)
_que.push(tmp->left);
if(tmp->right != nullptr)
_que.push(tmp->right);
}
++count;
}
return vv;
}
};
标签：predecessor,right,中序,nullptr,vector,二叉树,push,root,前序 来源： https://blog.csdn.net/w903414/article/details/112193240