[Swift]LeetCode285. 二叉搜索树中的中序后继节点 $ Inorder Successor in BST

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Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

The successor of a node p is the node with the smallest key greater than p.val.

Example 1:

Input: root = [2,1,3], p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of TreeNode type.

Example 2:

Input: root = [5,3,6,2,4,null,null,1], p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is null.

Note:

1. If the given node has no in-order successor in the tree, return null.
2. It's guaranteed that the values of the tree are unique.

---

给定一个二进制搜索树及其节点，在BST中查找该节点的顺序继承者。

 节点p的后续节点是键最小大于p.val的节点。

例 1:

输入: root = [2,1,3], p = 1
输出: 2
说明：1的顺序继承节点为2。请注意，p和返回值都是treenode类型。

例  2:

输入: root = [5,3,6,2,4,null,null,1], p = 6
输出: null
说明：当前节点没有按顺序的后续节点，因此答案为空。

注：

1. 如果给定的节点在树中没有顺序继承者，则返回空。
2. 它保证了树的值是唯一的。

---

Solution:

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 inorderSuccessor(_ root: TreeNode?,_ p: TreeNode?) -> TreeNode? {
        var root = root
        var res:TreeNode? = nil
        while(root != nil)
        {
            if root!.val > p!.val
            {
                res = root
                root = root?.left
            }
            else
            {
                root = root?.right
            }
        }
        return res
    }
}