代码随想录算法训练营第36期DAY14

DAY14（周二）

过了。

/**
* 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:
void preorderfunction(TreeNode* cur,vector<int>&res)
{
if(cur==nullptr) return;
res.push_back(cur->val);
preorderfunction(cur->left,res);
preorderfunction(cur->right,res);
}
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
preorderfunction(root,res);
return res;
}
};

过了

/**
* 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:
void postorderfunction(TreeNode* cur,vector<int>&res)
{
if(cur==nullptr) return;
postorderfunction(cur->left,res);
postorderfunction(cur->right,res);
res.push_back(cur->val);
}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> res;
postorderfunction(root,res);
return res;
}
};

/**
* 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:
void inorderfunction(TreeNode* cur,vector<int> &res)
{
if(cur==nullptr) return;
inorderfunction(cur->left,res);
res.push_back(cur->val);
inorderfunction(cur->right,res);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
inorderfunction(root,res);
return res;
}
};

这里都需要二刷，来检验和加深印象。

图片来自代码随想录

/**
* 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) {
vector<int> res;
stack<TreeNode*> stk;
if(root==nullptr) return res;
stk.push(root);
while(!stk.empty())
{
//先获取，以便进行迭代
TreeNode* cur=stk.top();
stk.pop();
res.push_back(cur->val);
//空结点不入栈，记住这里是要入栈，而不是单纯的更新cur
if(cur->right) stk.push(cur->right);
if(cur->left) stk.push(cur->left);
}
return res;
}
};

/**
* 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) {
vector<int> res;
stack<TreeNode*> stk;
TreeNode* cur=root;
if(root==nullptr) return res;
while(cur!=NULL||!stk.empty())
{
if(cur!=NULL)//根据先进后出的栈特点以及，中序遍历的特点：访问顺序与输出顺序相反，来入栈
{
//访问过的都入栈
stk.push(cur);
cur=cur->left;//找最左的孩子
}else{
cur=stk.top();//要处理它
stk.pop();
res.push_back(cur->val);//中。（自己是最左的，那么自己就没有左孩子了，自己就是中了）。
//那么最左孩子的头上：中节点怎么办呢？不用着急，下一次循环会处理，因为有!stk.empty();
cur=cur->right; //右
}
}
return res;
}
};

/**
* 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) {
vector<int> res;
stack<TreeNode*> stk;
if(root==nullptr) return res;
stk.push(root);
while(!stk.empty())
{
//先获取，以便进行迭代
TreeNode* cur=stk.top();
stk.pop();
res.push_back(cur->val);
//空结点不入栈，记住这里是要入栈，而不是单纯的更新cur
if(cur->left) stk.push(cur->left);
if(cur->right) stk.push(cur->right);
}
reverse(res.begin(),res.end());
return res;
}
};