Merge pull request Hearen#103 from cxxly/chenxiaoxu

Chenxiaoxu

Author: Hearen

ChenXiaoxu/101. Symmetric Tree.md

@@ -0,0 +1,96 @@
+# 101. Symmetric Tree
+
+## Problem
+
+Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
+
+For example, this binary tree is symmetric:
+```
+    1
+   / \
+  2   2
+ / \ / \
+3  4 4  3
+But the following is not:
+    1
+   / \
+  2   2
+   \   \
+   3    3
+```
+Note:
+Bonus points if you could solve it both recursively and iteratively.
+
+tag:
+
+## Solution
+
+### 递归
+
+树中两个点对称(n1,n2)， 递归判断n1的左子树与n2的右子树，n2的右子树与n2的左子树是否对称
+
+**java**
+```java
+    public boolean isSymmetric(TreeNode root) {
+       return root==null || helper(root.left, root.right); 
+    }
+    
+    boolean helper(TreeNode n1, TreeNode n1) {
+        if(n1==null || n2==null) return n1==n2;
+        if(n1.val != n2.val) return false;
+        return helper(n1.left, n2.right)&&helper(n1.right, n2.left);
+    }
+```
+
+**go**
+```go
+
+```
+
+### 迭代
+
+**模拟递归工作栈**
+
+```java
+    public boolean isSymmetric(TreeNode root) {
+       if(root==null) return true;
+       
+       Stack<TreeNode> s = new Stack<TreeNode>();
+       
+       if(root.left!=null && root.right!=null) {
+           s.push(root.left);
+           s.push(root.right);
+       } else if(root.left==null && root.right==null) {
+           return true;
+       } else {
+           return false;
+       }
+           
+       TreeNode left, right;
+       
+       while(!s.isEmpty()) {
+          // if(s.size()%2!=0) return false;
+           right = s.pop();
+           left = s.pop();
+           if(right.val!=left.val) return false;
+           
+           if(left.left!=null && right.right!=null) {
+               s.push(left.left);
+               s.push(right.right);
+           } else if(!(left.left==null&&right.right==null)) {
+              return false;
+           }
+            
+           if(left.right!=null && right.left!=null) {
+               s.push(left.right);
+               s.push(right.left);
+           } else if(!(left.right==null && right.left==null)) {
+               return false;
+           }
+       }
+       
+       return true;
+    }
+```
+
+**层次遍历，判断每层是否对称**

ChenXiaoxu/226. Invert Binary Tree.md

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+# 226. Invert Binary Tree
+
+## Problem
+nvert a binary tree.
+
+     4
+   /   \
+  2     7
+ / \   / \
+1   3 6   9
+to
+     4
+   /   \
+  7     2
+ / \   / \
+9   6 3   1
+
+tag:
+
+## Solution
+
+>如果节点为空，返回
+>否则
+- 递归转换左子树
+- 递归转换右子树
+- 对换左右子树
+
+**java**
+```java
+    public TreeNode invertTree(TreeNode root) {
+        if(root==null) return root;
+        TreeNode tmp = root.left;
+        root.left = invertTree(root.right);
+        root.right = invertTree(tmp);
+        return root;
+    }
+```
+
+**go**
+```go
+
+```

ChenXiaoxu/257. Binary Tree Paths.md

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+# 257. Binary Tree Paths
+
+## Problem
+
+Given a binary tree, return all root-to-leaf paths.
+
+For example, given the following binary tree:
+
+   1
+ /   \
+2     3
+ \
+  5
+All root-to-leaf paths are:
+
+["1->2->5", "1->3"]
+
+tag:
+- Tree
+
+## Solution
+
+**java**
+```java
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> res = new ArrayList<String>();
+        helper(res, new StringBuilder(), root);
+        return res;
+    }
+    
+    void helper(List<String> res, StringBuilder str, TreeNode root) {
+        if(root==null) return;
+        int len = str.length();
+        str.append(root.val);
+        if(root.left==null && root.right==null) {
+            res.add(str.toString());
+        } else {
+            str.append("->");
+            helper(res, str, root.left);
+            helper(res, str, root.right);
+        }
+        str.setLength(len);
+    }
+```
+
+**go**
+```go
+
+```