Merge pull request Hearen#104 from cxxly/chenxiaoxu

Chenxiaoxu

Author: Hearen

ChenXiaoxu/110. Balanced Binary Tree.md

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+# 110. Balanced Binary Tree
+
+## Problem
+Given a binary tree, determine if it is height-balanced.
+
+For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
+
+tag:
+- tree
+
+## Solution
+根据平衡二叉树定义：
+> - 左右子树高度之差不大于1
+> - 左右子树也是平衡二叉树
+
+**java**
+```java
+    public boolean isBalanced(TreeNode root) {
+        if(root==null) return true;
+        return Math.abs(depth(root.left)-depth(root.right))<=1 && isBalanced(root.left) && isBalanced(root.right);
+    }
+    
+    int depth(TreeNode root) {
+        if(root==null) return 0;
+        return Math.max(depth(root.left),depth(root.right))+1;
+    }
+```
+
+**go**
+```go
+
+```

ChenXiaoxu/2. Add Two Numbers.md

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+# 2. Add Two Numbers
+
+## Problem
+You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
+
+Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 0 -> 8
+
+tag:
+- linked list
+
+
+## Solution
+链表模拟加法操作，高位补零简化代码
+**java**
+```java
+	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
+		int carry = 0, res = 0;
+		ListNode dummy = new ListNode(0);
+		ListNode p = dummy;
+		while (l1 != null || l2 != null || carry!=0) {
+			res = ((l1==null)?0:l1.val) + ((l2==null)?0:l2.val) + carry;
+			carry = res / 10;
+			p.next = new ListNode(res % 10);
+			p = p.next;
+			l1 = (l1==null)?l1:l1.next;
+			l2 = (l2==null)?l2:l2.next;
+		}
+		return dummy.next;
+	}
+```
+
+**go**
+```go
+
+```

ChenXiaoxu/98. Validate Binary Search Tree.md

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+# 98. Validate Binary Search Tree
+
+## Problem
+Given a binary tree, determine if it is a valid binary search tree (BST).
+
+Assume a BST is defined as follows:
+
+The left subtree of a node contains only nodes with keys less than the node's key.
+The right subtree of a node contains only nodes with keys greater than the node's key.
+Both the left and right subtrees must also be binary search trees.
+
+tag:
+
+## Solution
+中序遍历，然后判断是否为升序
+
+**java**
+```java
+    public boolean isValidBST(TreeNode root) {
+        List<Integer> res = new ArrayList();
+        inOrderTraverse(root, res);
+        for(int i=1; i<res.size(); i++) {
+            if(res.get(i)<=res.get(i-1)) return false;
+        }
+        return true;
+    }
+    
+    void inOrderTraverse(TreeNode root, List<Integer> res) {
+        if(root!=null){
+            inOrderTraverse(root.left, res);
+            res.add(root.val);
+            inOrderTraverse(root.right, res);
+        }
+    }
+```
+
+**go**
+```go
+
+```

ChenXiaoxu/conclusion/tree/Binary Search Tree.md

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+#  二叉搜索树
+
+## 定义
+
+对于树中的每个节点，左子树值不大于该节点，右子树值不小于该节点
+左右子树均为二叉搜索树
+
+## 操作
+- 判断是否是BST
+
+## 相关题目

ChenXiaoxu/conclusion/tree/balanced binary tree.md

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+#  平衡二叉树
+
+## 定义
+每个节点的两个字树深度相差不超过1
+
+## 操作
+- 判断是否是平衡二叉树
+
+## 相关题目

ChenXiaoxu/template - 副本 (3).md

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+# Title
+
+## Problem
+
+tag:
+
+## Solution
+
+**java**
+```java
+
+```
+
+**go**
+```go
+
+```

ChenXiaoxu/template - 副本 (4).md

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+# Title
+
+## Problem
+
+tag:
+
+## Solution
+
+**java**
+```java
+
+```
+
+**go**
+```go
+
+```