101. 对称二叉树
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if(!root) return true;
stack<TreeNode*> s;
s.push(root->left);
s.push(root->right);
while(!s.empty()){
auto p = s.top();
s.pop();
auto q = s.top();
s.pop();
if(!q&&!p) continue;
if(!p||!q) return false;
if(p->val!=q->val) return false;
s.push(p->left);
s.push(q->right);
s.push(p->right);
s.push(q->left);
}
return true;
}
};
102. 二叉树的层次遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
TreeNode* p=root;
vector<vector<int>> a;
queue<TreeNode*> q;
q.push(p);
if(!p)
return a;
while(!q.empty())
{
int size=q.size();
vector<int> b;
while(size--)
{
p=q.front();
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
b.push_back(p->val);
q.pop();
}
a.push_back(b);
}
return a;
}
};
103. 二叉树的锯齿形层次遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
queue<TreeNode*> q;
stack<TreeNode*> s;
vector<int> m;
vector<vector<int>> n;
int size;
int p=0;
if(!root)
return n;
q.push(root);
while(!q.empty())
{
TreeNode*a;
size=q.size();
while(size--)
{
a=q.front();
if(a->left)
q.push(a->left);
if(a->right)
q.push(a->right);
m.push_back(a->val);
q.pop();
}
p++;
if(p%2==0)
reverse(m.begin(),m.end());
n.push_back(m);
m.clear();
}
return n;
}
};104. 二叉树的最大深度
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxDepth(TreeNode* root) {
TreeNode* p=root;
vector<vector<int>> a;
queue<TreeNode*> q;
q.push(p);
if(!p)
return 0;
while(!q.empty())
{
int size=q.size();
vector<int> b;
while(size--)
{
p=q.front();
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
b.push_back(p->val);
q.pop();
}
a.push_back(b);
}
int kaka=a.size();
return kaka;
}
};107. 二叉树的层次遍历 II
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
TreeNode* p=root;
vector<vector<int>> a;
queue<TreeNode*> q;
q.push(p);
if(!p)
return a;
while(!q.empty())
{
int size=q.size();
vector<int> b;
while(size--)
{
p=q.front();
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
b.push_back(p->val);
q.pop();
}
a.push_back(b);
}
reverse(a.begin(),a.end());
return a;
}
};136. 只出现一次的数字
class Solution {
public:
int singleNumber(vector<int>& nums) {
map<int, int> m;
for (int i = 0; i < nums.size(); i++) {
// key 为数组中的值 value为数组中的值出现的次数
if (!m.count(nums[i])) {
m[nums[i]] = 1;
}else {
m[nums[i]] += 1;
}
}
for (int i = 0; i < nums.size(); i++) {
if (m[nums[i]] == 1) {
return nums[i];
}
}
return -1;
}
};
144. 二叉树的前序遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
TreeNode*h= root;
vector<int> q;
stack<TreeNode*> a;
int k;
TreeNode*p;
if(!root)
return q;
a.push(h);
while(!a.empty())
{
p=a.top();
a.pop();
q.push_back(p->val);
if(p->right)
a.push(p->right);
if(p->left)
a.push(p->left);
}
return q;
}
};145. 二叉树的后序遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
TreeNode*h= root;
vector<int> q;
stack<TreeNode*> a;
int k;
TreeNode*p;
if(!root)
return q;
a.push(h);
while(!a.empty())
{
p=a.top();
if(p->left)
{
a.push(p->left);
p->left=NULL;
}
else
{
if(p->right)
{
a.push(p->right);
p->right=NULL;
}
else
{
q.push_back(p->val);
a.pop();
}
}
}
return q;
}
};150. 逆波兰表达式求值
class Solution {
public:
int evalRPN(vector<string>& tokens) {
stack<int> s_num;
int num1,num2;
for(auto token:tokens){
if(token=="+"||token=="-"||token=="*"||token=="/"){
num2 = s_num.top();
s_num.pop();
num1 = s_num.top();
s_num.pop();
char c = token[0];
switch(c){
case '+':
s_num.push(num1+num2);
break;
case '-':
s_num.push(num1-num2);
break;
case '*':
s_num.push(num1*num2);
break;
case '/':
s_num.push(num1/num2);
break;
}
}
else{
s_num.push(atoi(token.c_str()));
}
}
return s_num.top();
}
};