[Swift]LeetCode1038. 从二叉搜索树到更大和树 | 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.

---

给出二叉搜索树的根节点，该二叉树的节点值各不相同，修改二叉树，使每个节点 node 的新值等于原树的值之和，这个值应该大于或等于 node.val。

提醒一下，二叉搜索树满足下列约束条件：

• 节点的左子树仅包含键小于节点键的节点。
• 节点的右子树仅包含键大于节点键的节点。
• 左右子树也必须是二叉搜索树。

示例：

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

提示：

1. 树中的节点数介于 1 和 100 之间。
2. 每个节点的值介于 0 和 100 之间。
3. 给定的树为二叉搜索树。

---

Runtime: 8 ms

Memory Usage: 19.2 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func bstToGst(_ root: TreeNode?) -> TreeNode? {
        var root = root
        var acc:Int = 0
        dfs(&root,&acc)
        return root        
    }
    
    func dfs(_ node:inout TreeNode?,_ acc:inout Int)
    {
        if node != nil
        {
            dfs(&node!.right, &acc)
            var temp:Int = node!.val
            node!.val += acc
            acc += temp
            dfs(&node!.left, &acc)
        }        
    }
}

---

8ms 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
     var sum = 0;
    func bstToGst(_ root: TreeNode?) -> TreeNode? {
        
        guard let rootd = root else {return nil}
        bstToGst(rootd.right);
        rootd.val += sum;
        sum = rootd.val;
        bstToGst(rootd.left);
        return rootd
        
    }
}