feat: add 101

Author: Blankj

note/001/README.md

@@ -38,7 +38,6 @@ class Solution {
 }
 ```
 
-
 ## 思路1
 
 利用HashMap作为存储，键为目标值减去当前元素值，索引为值，比如`i = 0`时，此时首先要判断`nums[0] = 2`是否在map中，如果不存在，那么插入键值对`key = 9 - 2 = 7, value = 0`，之后当`i = 1`时，此时判断`nums[1] = 7`已存在于map中，那么取出该`value = 0`作为第一个返回值，当前`i`作为第二个返回值，具体代码如下所示。

note/100/README.md

@@ -11,7 +11,7 @@ Two binary trees are considered equal if they are structurally identical and the
 
 ## 思路
 
-题意是
+题意是比较两棵二叉树是否相同，那么我们就深搜比较各个节点即可。
 
 ``` java
 /**

note/101/README.md

@@ -0,0 +1,106 @@
+# [Symmetric Tree][title]
+
+## Description
+
+Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
+
+For example, this binary tree `[1,2,2,3,4,4,3]` is symmetric:
+
+```
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+
+```
+
+But the following `[1,2,2,null,3,null,3]` is not:
+
+```
+    1
+   / \
+  2   2
+   \   \
+   3    3
+
+```
+
+**Note:**
+Bonus points if you could solve it both recursively and iteratively.
+
+**Tags:** Tree, Depth-first Search, Breadth-first Search
+
+
+## 思路0
+
+题意是判断一棵二叉树是否左右对称，首先想到的是深搜，比较根结点的左右两棵子树是否对称，如果左右子树的值相同，那么再分别对左子树的左节点和右子树的右节点，左子树的右节点和右子树的左节点做比较即可。
+
+``` java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public boolean isSymmetric(TreeNode root) {
+        return root == null || isSymmetricHelper(root.left, root.right);
+    }
+
+    public boolean isSymmetricHelper(TreeNode left, TreeNode right) {
+        if (left == null || right == null) return left == right;
+        if (left.val != right.val) return false;
+        return isSymmetricHelper(left.left, right.right) && isSymmetricHelper(left.right, right.left);
+    }
+}
+```
+
+## 思路1
+
+第二种思路就是宽搜了，宽搜肯定要用到队列，Java中可用`LinkedList`替代，也是要做到左子树的左节点和右子树的右节点，左子树的右节点和右子树的左节点做比较即可。
+
+``` java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+     public boolean isSymmetric(TreeNode root) {
+        if (root == null) return true;
+        LinkedList<TreeNode> q = new LinkedList<>();
+        q.add(root.left);
+        q.add(root.right);
+        TreeNode left, right;
+        while (q.size() > 1) {
+            left = q.pop();
+            right = q.pop();
+            if (left == null && right == null) continue;
+            if (left == null || right == null) return false;
+            if (left.val != right.val) return false;
+            q.add(left.left);
+            q.add(right.right);
+            q.add(left.right);
+            q.add(right.left);
+        }
+        return true;
+    }
+}
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/symmetric-tree
+[ajl]: https://github.com/Blankj/awesome-java-leetcode