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Merge pull request gzc426#20 from gzc426/master
同步更新乔戈里leetcode每日一题
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‎2018.11.27-leetcode125/。。。.md

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```
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public boolean isPalindrome(String s){
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if (s.length() >= 2){
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int left = 0,right = s.length()-1;
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}
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return true;
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}
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```

‎2018.11.28-leetcode151/。。。.md

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```
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public String reverseWords(String s) {
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int i = s.length()-1,wordRight=-1;
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StringBuilder builder = new StringBuilder();
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}
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return builder.toString();
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}
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```

‎2018.11.29-leetcode443/。。。.md

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public int compress(char[] chars) {
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```
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public int compress(char[] chars) {
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int cur = 1;//压缩后字符串的长度
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int repet = 1;
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for (int i = 1; i < chars.length; i++){
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Log.d(new String(chars));
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return cur;
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}
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```

‎2018.12.04-leetcode101/Istoryforever.md

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leetcode of 101
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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* p1,TreeNode * p2){
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if((p1 == nullptr && p2!=nullptr) || (p2 == nullptr && p1!=nullptr)) return false;
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if(p1 == nullptr && p2 == nullptr) return true;
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if(p1->val != p2->val) return false;
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return isSymmetric(p1->left,p2->right) && isSymmetric(p1->right,p2->left);
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}
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bool isSymmetric(TreeNode* root) {
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return isSymmetric(root,root);
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}
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};
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```

‎2018.12.04-leetcode101/sourcema.md

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# LeetCode 101
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/**
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* Definition for a binary tree node.
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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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* }
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*/
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class Solution {
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public boolean 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 symmetric(root.left, root.right);
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}
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private boolean symmetric(TreeNode p, TreeNode q) {
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if (p == null && q == null) {
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return true;
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}
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if ((p == null && q != null) || (p != null && q == null) || p.val != q.val) {
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return false;
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}
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return symmetric(p.left, q.right) && symmetric(p.right, q.left);
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}
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}

‎2018.12.04-leetcode101/啦啦啦.md

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```java
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/**
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* Definition for a binary tree node.
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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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* }
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*/
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class Solution {
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public boolean isSymmetric(TreeNode root) {
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return cal(root,root);
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}
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boolean cal(TreeNode r1,TreeNode r2){
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if(r1==null&&r2==null)return true;
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if(r1==null||r2==null)return false;
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return r1.val==r2.val&&cal(r1.left,r2.right)&&cal(r1.right,r2.left);
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}
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}
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```

‎2018.12.04-leetcode101/李碧涵.md

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### [101. Symmetric Tree](https://leetcode.com/problems/symmetric-tree/description/)
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**题目描述**
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> 判断是否是镜像二叉树（对称）
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**例子**
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略
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**思想**
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（递归）
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定义一辅助函数，输入为left和right，判断是否相等。
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（迭代）
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定义两个栈，层次遍历时判断，并分别从左向右和从右向左存储。
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**解法1**
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递归
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```python
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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def isSymmetric(self, root):
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"""
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:type root: TreeNode
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:rtype: bool
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"""
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if not root:
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return True
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return self.helper(root.left, root.right)
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def helper(self, left, right):
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if not left and not right:
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return True
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if not left or not right or left.val != right.val:
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return False
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return self.helper(left.left, right.right) and self.helper(left.right, right.left)
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```
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**解法2**
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迭代
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```python
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# Definition for a binary tree node.
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# class TreeNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution(object):
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def isSymmetric(self, root):
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"""
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:type root: TreeNode
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:rtype: bool
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"""
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if not root or not root.left and not root.right:
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return True
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if not root.left or not root.right:
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return False
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queue1, queue2 = [root.left], [root.right]
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while queue1 and queue2:
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node1 = queue1.pop(0)
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node2 = queue2.pop(0)
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if node1.val != node2.val:
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return False
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if node1.left and node2.right:
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queue1.append(node1.left)
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queue2.append(node2.right)
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elif node1.left or node2.right:
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return False
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if node1.right and node2.left:
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queue1.append(node1.right)
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queue2.append(node2.left)
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elif node1.right or node2.left:
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return False
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return True
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```

‎2018.12.04-leetcode101/杨.md

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```
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/**
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* 如果一个树的左子树与右子树镜像对称，那么这个树是对称的。
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* 如果同时满足下面的条件，两个树互为镜像：
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* 1.它们的两个根结点具有相同的值。
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* 2.每个树的右子树都与另一个树的左子树镜像对称。
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*/
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public boolean 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 isMirrorSymmetry(root.left,root.right);
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}
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public boolean isMirrorSymmetry(TreeNode left,TreeNode right){
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if(left==null && right==null){
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return true;
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}
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if(left==null || right==null){
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return false;
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}
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return left.val==right.val && isMirrorSymmetry(left.left,right.right) && isMirrorSymmetry(left.right,right.left);
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}
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```

‎2018.12.05-leetcode102/GatesMa.md

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# C++
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```cpp
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class Solution {
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public:
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vector<vector<int>> res;
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void check(TreeNode* root, int depth){
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if(root == NULL) return ;
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if(res.size() == depth)
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{
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res.push_back(vector<int>());
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}
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res[depth].push_back(root->val);
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check(root->left, depth + 1);
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check(root->right, depth + 1);
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}
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vector<vector<int>> levelOrder(TreeNode* root) {
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check(root, 0);
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return res;
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}
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};
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```

‎2018.12.05-leetcode102/Istoryforever.md

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leetcode of 102
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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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struct DataNode{
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TreeNode * p;
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int depth;
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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>> ans;
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if(root == nullptr) return ans;
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queue<DataNode> q;
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int depth = 0;
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q.push(DataNode{root,depth});
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vector<int>t;
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while(!q.empty()){
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DataNode dn = q.front();
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q.pop();
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if(depth == dn.depth){
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t.push_back(dn.p->val);
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}else{
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depth = dn.depth;
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ans.push_back(t);
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t.clear();
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t.push_back(dn.p->val);
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}
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if(dn.p->left != nullptr)
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q.push(DataNode{dn.p->left,depth+1});
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if(dn.p->right != nullptr)
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q.push(DataNode{dn.p->right,depth+1});
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}
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ans.push_back(t);
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return ans;
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}
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};
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```

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