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‎2018.11.30-leetcode890/神秘的火柴人.md

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class Solution {
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public:
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vector<string> findAndReplacePattern(vector<string>& words, string pattern) {
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vector<string> ret;
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int size = words.size();
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int n = pattern.length();
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for(int i=0; i<size; i++)
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{
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map<char, char> m;
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map<char, char> m1;
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int j = 0;
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for(j=0; j<n; j++)
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{
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if(m.find(pattern[j])==m.end())
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{
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m[pattern[j]] = words[i][j];
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}
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if(m1.find(words[i][j])==m1.end())
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{
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m1[words[i][j]] = pattern[j];
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}
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if(m.find(pattern[j])!=m.end() && m[pattern[j]] != words[i][j])
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break;
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if(m1.find(words[i][j])!=m.end() && m1[words[i][j]]!=pattern[j])
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break;
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}
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if(j == n)
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ret.push_back(words[i]);
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}
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return ret;
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}
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};

‎2018.12.04-leetcode101/MQQM.cpp

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/*
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题目：
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给定一个二叉树，检查它是否是镜像对称的。
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做法：
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给定对应位置上的两个节点，递归判断这两个节点是否一致。
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参考：
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https://www.cnblogs.com/xiaozhuyang/p/7376749.html
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*/
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/**
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* Definition for a binary tree node.
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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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class Solution {
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public:
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bool isSymmetric(TreeNode* root) {
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if( root == NULL ){
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return true;
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}
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return isEqual(root->left, root->right);
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}
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bool isEqual(TreeNode* left, TreeNode* right){//给定的两个节点已经是对应位置上的节点
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if( left == NULL && right == NULL ){//两个节点都是空
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return true;
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}
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if( left == NULL || right == NULL ){//有一个节点是空
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return false;
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}
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if( left->val != right->val ){//两个节点都存在，但是不相同
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return false;
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}
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return isEqual(left->left, right->right) && isEqual(left->right, right->left);
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}
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};

‎2018.12.04-leetcode101/神秘的火柴人.md

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```
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/**
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* Definition for a binary tree node.
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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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class Solution {
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public:
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bool isSymmetric(TreeNode* root) {
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if(!root) return true;
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queue<TreeNode*> q1;
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queue<TreeNode*> q2;
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q1.push(root->left);
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q2.push(root->right);
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while(!q1.empty() && !q2.empty())
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{
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TreeNode *l = q1.front();
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TreeNode *r = q2.front();
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q1.pop();
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q2.pop();
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if( (!l&&r) || (l&&!r) ) return false;
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if(l && r)
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{
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if(l->val != r->val)
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return false;
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q1.push(l->left);
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q1.push(l->right);
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q2.push(r->right);
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q2.push(r->left);
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}
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}
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return true;
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}
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};
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```

‎2018.12.05-leetcode102/MQQM.cpp

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/*
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题目：
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给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
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做法：用队列。
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*/
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/**
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* Definition for a binary tree node.
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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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class Solution {
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public:
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vector<vector<int>> levelOrder(TreeNode* root) {
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vector<vector<int>> rst;
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if(root == NULL){
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return rst;
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}
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queue<TreeNode*> que;
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que.push(root);
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while( !que.empty() ){
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int len=que.size();
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vector<int> levelvec;
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for(int i=0; i<len; i++){
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TreeNode* p=que.front();
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que.pop();
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levelvec.push_back(p->val);
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if(p->left != NULL){
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que.push(p->left);
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}
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if(p->right != NULL){
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que.push(p->right);
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}
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}
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rst.push_back(levelvec);
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}
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return rst;
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}
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};

‎2018.12.05-leetcode102/kaizen

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/**
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* @author kaizen
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* @since 2018-12-07
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**/
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public class 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; }
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class Solution {
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public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
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if(root == null)
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return new ArrayList<>();
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List<List<Integer>> res = new ArrayList<>();
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Queue<TreeNode> queue = new LinkedList<TreeNode>();
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queue.add(root);
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while(!queue.isEmpty()){
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int count = queue.size();
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List<Integer> list = new ArrayList<Integer>();
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while(count > 0){
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TreeNode node = queue.poll();
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list.add(node.val);
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if(node.left != null)
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queue.add(node.left);
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if(node.right != null)
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queue.add(node.right);
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count--;
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}
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res.add(list);
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}
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return res;
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}
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}
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}

‎2018.12.05-leetcode102/神秘的火柴人.md

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```
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/**
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* Definition for a binary tree node.
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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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class Solution {
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public:
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vector<vector<int>> levelOrder(TreeNode* root) {
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vector<vector<int>> ret;
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if(!root) return ret;
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queue<TreeNode*> q;
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q.push(root);
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while(!q.empty())
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{
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int n = q.size();
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vector<int> tmp;
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while(n>0)
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{
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TreeNode *node = q.front();
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q.pop();
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n--;
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tmp.push_back(node->val);
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if(node->left)
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q.push(node->left);
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if(node->right)
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q.push(node->right);
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}
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ret.push_back(tmp);
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}
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return ret;
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}
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};
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```

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