285. Inorder Successor in BST

Given the root of a binary search tree and a node p in it, return the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, return null.

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.

思路Coding化:

class Solution:
    def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> Optional[TreeNode]:
        successor = None

        #不斷往下找，當root = None停止
        while root:
            #當p大於root，代表只能往右邊的樹尋找
            if root.val <= p.val:
                root = root.right
            #當p小於root，代表答案可能在左邊的樹，root此時可能為答案，因此更新successor
            #注意這題是要回傳node型態
            else:
                successor = root
                root = root.left
        return successor

        #O(n)
        #O(1)

Time Complexity: O(n)

Space Complexity: O(1)