leetcode update

content/posts/leetcode.md

@@ -7982,3 +7982,56 @@ func minKBitFlips(nums []int, k int) int {
     return -1
 }
 ```
+
+## day120 2024-06-25
+### 1038. Binary Search Tree to Greater Sum Tree 
+
+Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.
+
+As a reminder, a binary search tree is a tree that satisfies these constraints:
+
+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.
+
+![0625JppqmJoxVzA1](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0625JppqmJoxVzA1.png) 
+
+### 题解
+
+考虑到树为一棵二叉搜索树, 因此对于任何一个节点, 比它大的节点有其右子树的节点, 若其父节点为祖父节点的左子节点, 则还有其祖父节点以及祖父节点的右子树的节点. 以此类推. 考虑最简单的仅有三个节点的情况: 一个父节点和左右两个子节点. 对于左子节点而言, 需要加父节点和右子节点的值. 对于父节点要加右子节点的值. 右子节点不用处理. 因此通过对树遍历将问题转化为类似三个节点的情况. 左右子树均可视为一个节点. 从根节点开始, 对节点进行如下操作, 递归遍历右子树, 计算右子树所有节点的和返回. 将当前节点的值加上右子树的和, 将和传递给左子树并递归遍历左子树, 返回左子树所有节点的和. 最终返回根节点和左右两个子树的和. 注意更新节点值和计算子树和是分开的, 更新节点值要将传递的值和当前值以及右子树的和相加. 但返回时返回的是原来的节点值对应的子树和.  
+
+### 代码
+```go 
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func bstToGst(root *TreeNode) *TreeNode {
+    rightFirst(root, 0)
+    return root
+}
+
+func rightFirst(root *TreeNode, pass int)int{
+    temp := root.Val
+    if root.Left == nil && root.Right == nil{
+        root.Val += pass
+        return temp
+    }
+    rightSum := 0
+    if root.Right != nil{
+        rightSum = rightFirst(root.Right, pass)
+    }
+    leftSum := 0
+    if root.Left != nil{
+        leftSum = rightFirst(root.Left, rightSum+temp+pass)
+    }
+    root.Val = root.Val + pass + rightSum
+    return temp + rightSum + leftSum
+}
+
+```
+