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README.md
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README.md
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# 算法模板
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## 二分查找法
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```
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class Solution {
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public:
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int searchInsert(vector<int>& nums, int target) {
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int n = nums.size();
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int left = 0;
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int right = n; // 我们定义target在左闭右开的区间里，[left, right)
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while (left < right) { // 因为left == right的时候，在[left, right)是无效的空间
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int middle = left + ((right - left) >> 1);
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if (nums[middle] > target) {
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right = middle; // target 在左区间，因为是左闭右开的区间，nums[middle]一定不是我们的目标值，所以right = middle，在[left, middle)中继续寻找目标值
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} else if (nums[middle] < target) {
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left = middle + 1; // target 在右区间，在 [middle+1, right)中
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} else { // nums[middle] == target
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return middle; // 数组中找到目标值的情况，直接返回下标
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}
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}
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return right;
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}
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};
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```
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## KMP
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```
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void kmp(int* next, const string& s){
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next[0] = -1;
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int j = -1;
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for(int i = 1; i < s.size(); i++){
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while (j >= 0 && s[i] != s[j + 1]) {
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j = next[j];
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}
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if (s[i] == s[j + 1]) {
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j++;
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}
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next[i] = j;
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}
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}
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```
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## 二叉树
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二叉树的定义：
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```
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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};
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```
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### 深度优先遍历（递归）
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前序遍历（中左右）
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```
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void traversal(TreeNode* cur, vector<int>& vec) {
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if (cur == NULL) return;
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vec.push_back(cur->val); // 中 ，同时也是处理节点逻辑的地方
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traversal(cur->left, vec); // 左
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traversal(cur->right, vec); // 右
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}
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```
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中序遍历（左中右）
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```
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void traversal(TreeNode* cur, vector<int>& vec) {
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if (cur == NULL) return;
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traversal(cur->left, vec); // 左
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vec.push_back(cur->val); // 中 ，同时也是处理节点逻辑的地方
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traversal(cur->right, vec); // 右
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}
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```
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中序遍历（中左右）
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```
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void traversal(TreeNode* cur, vector<int>& vec) {
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if (cur == NULL) return;
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vec.push_back(cur->val); // 中 ，同时也是处理节点逻辑的地方
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traversal(cur->left, vec); // 左
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traversal(cur->right, vec); // 右
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}
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```
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### 深度优先遍历（迭代法）
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相关题解：[0094.二叉树的中序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0094.二叉树的中序遍历.md)
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前序遍历（中左右）
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```
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vector<int> preorderTraversal(TreeNode* root) {
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vector<int> result;
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stack<TreeNode*> st;
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if (root != NULL) st.push(root);
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while (!st.empty()) {
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TreeNode* node = st.top();
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if (node != NULL) {
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st.pop();
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if (node->right) st.push(node->right); // 右
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if (node->left) st.push(node->left); // 左
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st.push(node); // 中
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st.push(NULL);
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} else {
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st.pop();
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node = st.top();
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st.pop();
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result.push_back(node->val); // 节点处理逻辑
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}
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}
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return result;
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}
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```
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中序遍历（左中右）
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```
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vector<int> inorderTraversal(TreeNode* root) {
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vector<int> result; // 存放中序遍历的元素
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stack<TreeNode*> st;
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if (root != NULL) st.push(root);
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while (!st.empty()) {
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TreeNode* node = st.top();
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if (node != NULL) {
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st.pop();
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if (node->right) st.push(node->right); // 右
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st.push(node); // 中
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st.push(NULL);
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if (node->left) st.push(node->left); // 左
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} else {
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st.pop();
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node = st.top();
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st.pop();
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result.push_back(node->val); // 节点处理逻辑
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}
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}
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return result;
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}
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```
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后序遍历（左右中）
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```
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vector<int> postorderTraversal(TreeNode* root) {
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vector<int> result;
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stack<TreeNode*> st;
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if (root != NULL) st.push(root);
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while (!st.empty()) {
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TreeNode* node = st.top();
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if (node != NULL) {
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st.pop();
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st.push(node); // 中
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st.push(NULL);
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if (node->right) st.push(node->right); // 右
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if (node->left) st.push(node->left); // 左
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} else {
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st.pop();
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node = st.top();
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st.pop();
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result.push_back(node->val); // 节点处理逻辑
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}
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}
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return result;
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}
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```
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### 广度优先遍历（队列）
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相关题解：[0102.二叉树的层序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0102.二叉树的层序遍历.md)
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```
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vector<vector<int>> levelOrder(TreeNode* root) {
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queue<TreeNode*> que;
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if (root != NULL) que.push(root);
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vector<vector<int>> result;
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while (!que.empty()) {
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int size = que.size();
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vector<int> vec;
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for (int i = 0; i < size; i++) {// 这里一定要使用固定大小size，不要使用que.size()
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TreeNode* node = que.front();
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que.pop();
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vec.push_back(node->val); // 节点处理的逻辑
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if (node->left) que.push(node->left);
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if (node->right) que.push(node->right);
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}
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result.push_back(vec);
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}
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return result;
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}
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```
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可以直接解决如下题目：
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* [0102.二叉树的层序遍历](https://github.com/youngyangyang04/leetcode/blob/master/problems/0102.二叉树的层序遍历.md)
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* [0199.二叉树的右视图](https://github.com/youngyangyang04/leetcode/blob/master/problems/0199.二叉树的右视图.md)
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* [0637.二叉树的层平均值](https://github.com/youngyangyang04/leetcode/blob/master/problems/0637.二叉树的层平均值.md)
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* [0104.二叉树的最大深度 （迭代法）](https://github.com/youngyangyang04/leetcode/blob/master/problems/0104.二叉树的最大深度.md)
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* [0111.二叉树的最小深度（迭代法）]((https://github.com/youngyangyang04/leetcode/blob/master/problems/0111.二叉树的最小深度.md))
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* [0222.完全二叉树的节点个数（迭代法）](https://github.com/youngyangyang04/leetcode/blob/master/problems/0222.完全二叉树的节点个数.md)
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### 二叉树深度
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```
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int getDepth(TreeNode* node) {
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if (node == NULL) return 0;
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return 1 + max(getDepth(node->left), getDepth(node->right));
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}
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```
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### 二叉树节点数量
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```
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int countNodes(TreeNode* root) {
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if (root == NULL) return 0;
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return 1 + countNodes(root->left) + countNodes(root->right);
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}
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```
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## 回溯算法
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```
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backtracking() {
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if (终止条件) {
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存放结果;
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}
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for (选择：选择列表（可以想成树中节点孩子的数量）) {
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递归，处理节点;
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backtracking();
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回溯，撤销处理结果
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}
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}
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```
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（持续补充ing）
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[各类基础算法模板](https://github.com/youngyangyang04/leetcode/blob/master/problems/算法模板.md)
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# LeetCode 最强题解:
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@@ -469,6 +230,7 @@ backtracking() {
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|[0617.合并二叉树](https://github.com/youngyangyang04/leetcode/blob/master/problems/0617.合并二叉树.md) |树 |简单|**递归** **迭代**|
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|[0637.二叉树的层平均值](https://github.com/youngyangyang04/leetcode/blob/master/problems/0637.二叉树的层平均值.md) |树 |简单|**广度优先搜索/队列**|
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|[0654.最大二叉树](https://github.com/youngyangyang04/leetcode/blob/master/problems/0654.最大二叉树.md) |树 |中等|**递归**|
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|[0685.冗余连接II](https://github.com/youngyangyang04/leetcode/blob/master/problems/0685.冗余连接II.md) | 并查集/树/图 |困难|**并查集**|
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|[0700.二叉搜索树中的搜索](https://github.com/youngyangyang04/leetcode/blob/master/problems/0700.二叉搜索树中的搜索.md) |树 |简单|**递归** **迭代**|
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|[0701.二叉搜索树中的插入操作](https://github.com/youngyangyang04/leetcode/blob/master/problems/0701.二叉搜索树中的插入操作.md) |树 |简单|**递归** **迭代**|
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|[0705.设计哈希集合](https://github.com/youngyangyang04/leetcode/blob/master/problems/0705.设计哈希集合.md) |哈希表 |简单|**模拟**|
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