note

Author: guangxush

guangxu/202103/20210311.md

@@ -86,6 +86,10 @@ class Solution {
 
 ### Discuss
 
+这道题也是一道十分典型的动态规划的题目，我们可以将d个骰子看作一个骰子投掷了d次，那么，每投掷一次，都会和之前的投掷结果相关联。所以，我们可以定义一个二维数组dp，行表示投掷的次数，列表示投掷的数目之和，dp[i][j]表示一共投掷i次骰子时候所有数字之和为j的可能组合的数目。由于一个骰子只有f个面，因此需要求dp[i][j], 我们只需要遍历j前面的f个数据即可。
+
+![](assets/20210311-92b2f7b0.png)
+
 ## Review
 
 

guangxu/202103/20210312.md

@@ -1,11 +1,92 @@
 ## Algorithm
 
+[701. Insert into a Binary Search Tree](https://leetcode.com/problems/insert-into-a-binary-search-tree/)
+
 ### Description
 
+You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
+
+Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.
+
+
+Example 1:
+
+![](assets/20210312-bfd56585.png)
+
+```
+Input: root = [4,2,7,1,3], val = 5
+Output: [4,2,7,1,3,5]
+Explanation: Another accepted tree is:
+```
+
+![](assets/20210312-159a1bcd.png)
+
+Example 2:
+
+```
+Input: root = [40,20,60,10,30,50,70], val = 25
+Output: [40,20,60,10,30,50,70,null,null,25]
+```
+
+Example 3:
+
+```
+Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
+Output: [4,2,7,1,3,5]
+```
+
+Constraints:
+
+- The number of nodes in the tree will be in the range [0, 104].
+- -108 <= Node.val <= 108
+- All the values Node.val are unique.
+- -108 <= val <= 108
+- It's guaranteed that val does not exist in the original BST.
+
+
 ### Solution
 
-```java 
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public TreeNode insertIntoBST(TreeNode root, int val) {
+        if (root == null){
+            return new TreeNode(val);
+        }
+        helper(root, val);
+        return root;
+    }
 
+    public void helper(TreeNode root, int val){
+        if(root.left==null && val<root.val){
+            root.left = new TreeNode(val);
+            return;
+        }
+        if(root.right==null && val>root.val){
+            root.right = new TreeNode(val);
+            return;
+        }
+        if(val > root.val){
+            helper(root.right, val);
+        }else{
+            helper(root.left, val);
+        }
+    }
+}
 ```
 
 ### Discuss