春节刷题day11：LeetCode
剑指 Offer 27. 二叉树的镜像
剑指 Offer 55 - I. 二叉树的深度
剑指 Offer 54. 二叉搜索树的第k大节点
剑指 Offer 68 - I. 二叉搜索树的最近公共祖先
剑指 Offer 32 - II. 从上到下打印二叉树 II
剑指 Offer 55 - II. 平衡二叉树
剑指 Offer 28. 对称的二叉树
1、剑指 Offer 27. 二叉树的镜像
/**
* 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:
void slove(TreeNode* root){
if(root == NULL) return ;
swap(root -> left, root -> right);
mirrorTree(root -> left);
mirrorTree(root -> right);
}
TreeNode* mirrorTree(TreeNode* root) {
slove(root);
return root;
}
};
/**
* 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:
TreeNode* mirrorTree(TreeNode* root) {
stack<TreeNode*> st;
st.push(root);
while(!st.empty()){
TreeNode* now = st.top(); st.pop();
if(now == NULL) continue;
swap(now -> left, now -> right);
st.push(now -> left);
st.push(now -> right);
}
return root;
}
};
2、剑指 Offer 55 - I. 二叉树的深度
/**
* 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) {
if(root == NULL) return 0;
int l = maxDepth(root -> left);
int r = maxDepth(root -> right);
return max(l, r) + 1;
}
};
/**
* 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) {
int ans = -1;
stack<TreeNode*> st;
if(root == NULL) return 0;
root -> val = 1;
st.push(root);
while(!st.empty()){
TreeNode* now = st.top(); st.pop();
ans = max(ans, now -> val);
if(now -> left){
now -> left -> val = now -> val + 1;
st.push(now -> left);
}
if(now -> right){
now -> right -> val = now -> val + 1;
st.push(now -> right);
}
}
return ans;
}
};
3、剑指 Offer 54. 二叉搜索树的第k大节点
/**
* 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 kthLargest(TreeNode* root, int k) {
queue<TreeNode*> que; que.push(root);
vector<int> v;
while(!que.empty()){
TreeNode* now = que.front(); que.pop();
v.push_back(now -> val);
if(now -> left != NULL) que.push(now -> left);
if(now -> right != NULL) que.push(now -> right);
}
sort(v.begin(), v.end(), greater<int>());
for(int i = 0; i < v.size(); i++){
k--;
if(k == 0) return v[i];
}
return 0;
}
};
/**
* 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 ans = 0;
int k;
void slove(TreeNode* root){
if(root == NULL) return;
slove(root -> right);
if(--k == 0){
ans = root -> val; return ;
}
slove(root -> left);
}
int kthLargest(TreeNode* root, int k) {
this -> k = k;
slove(root);
return ans;
}
};
4、剑指 Offer 68 - I. 二叉搜索树的最近公共祖先
/**
* 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:
TreeNode* ans;
void slove(TreeNode* root, TreeNode* p, TreeNode* q){
if(root -> val > p -> val && root -> val > q -> val)
slove(root -> left, p, q);
else if(root -> val < p -> val && root -> val < q -> val)
slove(root -> right, p, q);
else{
ans = root; return ;
}
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
slove(root, p, q);
return ans;
}
};
/**
* 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:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
while(root){
if(root -> val < p -> val && root -> val < q -> val) root = root -> right;
else if(root -> val > p -> val && root -> val > q -> val) root = root -> left;
else break;
}
return root;
}
};
6、剑指 Offer 32 - II. 从上到下打印二叉树 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>> levelOrder(TreeNode* root) {
vector<vector<int>> ans;
if(root == NULL) return ans;
queue<TreeNode*> que; que.push(root);
while(!que.empty()){
int size = que.size();
vector<int> ANS;
while(size--){
TreeNode* now = que.front(); que.pop();
ANS.push_back(now -> val);
if(now -> left) que.push(now -> left);
if(now -> right) que.push(now -> right);
}
ans.push_back(ANS);
}
return ans;
}
};
7、剑指 Offer 55 - II. 平衡二叉树
/**
* 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:
int high(TreeNode* root){
if(root == NULL) return 0;
return max(high(root -> left), high(root -> right)) + 1;
}
bool isBalanced(TreeNode* root) {
if(root == NULL) return true;
int l = high(root -> left);
int r = high(root -> right);
if(abs(l - r) > 1) return false;
return isBalanced(root -> left) && isBalanced(root -> right);
}
};
8、剑指 Offer 28. 对称的二叉树
/**
* 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 slove(TreeNode* p, TreeNode* q){
if(p == NULL && q == NULL) return true;
else if(p == NULL && q != NULL) return false;
else if(p != NULL && q == NULL) return false;
else if(p -> val != q -> val) return false;
bool l = slove(p -> left, q -> right);
bool r = slove(p -> right, q -> left);
return l && r;
}
bool isSymmetric(TreeNode* root) {
if(root == NULL) return true;
return slove(root -> left, root -> right);
}
};
2021/2/16完结（写不动了，本来想着今天晚上再写几题的，可是看了会C++就懒得动了，欸！等着明天再写吧）。