[Swift]LeetCode1038. 從二叉搜索樹到更大和樹 | Binary Search Tree to Greater Sum Tree

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Given the root of a binary search tree with distinct values, modify it so that every node has a new value equal to the sum of the values of the original tree that are greater than or equal to node.val.

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.

Example 1:

Input: [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Note:

1. The number of nodes in the tree is between 1 and 100.
2. Each node will have value between 0 and 100.
3. The given tree is a binary search tree.

---

給出二叉搜索樹的根節點，該二叉樹的節點值各不相同，修改二叉樹，使每一個節點 node 的新值等於原樹的值之和，這個值應該大於或等於 node.val。

提醒一下，二叉搜索樹知足下列約束條件：

• 節點的左子樹僅包含鍵小於節點鍵的節點。
• 節點的右子樹僅包含鍵大於節點鍵的節點。
• 左右子樹也必須是二叉搜索樹。

示例：

輸入：[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
輸出：[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

提示：

1. 樹中的節點數介於 1 和 100 之間。
2. 每一個節點的值介於 0 和 100 之間。
3. 給定的樹爲二叉搜索樹。

---

Runtime: 8 ms

Memory Usage: 19.2 MB

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func bstToGst(_ root: TreeNode?) -> TreeNode? {
16         var root = root
17         var acc:Int = 0
18         dfs(&root,&acc)
19         return root        
20     }
21     
22     func dfs(_ node:inout TreeNode?,_ acc:inout Int)
23     {
24         if node != nil
25         {
26             dfs(&node!.right, &acc)
27             var temp:Int = node!.val
28             node!.val += acc
29             acc += temp
30             dfs(&node!.left, &acc)
31         }        
32     }
33 }

---

8ms 

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15      var sum = 0;
16     func bstToGst(_ root: TreeNode?) -> TreeNode? {
17         
18         guard let rootd = root else {return nil}
19         bstToGst(rootd.right);
20         rootd.val += sum;
21         sum = rootd.val;
22         bstToGst(rootd.left);
23         return rootd
24         
25     }
26 }