LeetCode 510. Inorder Successor in BST II

原题链接在这里：https://leetcode.com/problems/inorder-successor-in-bst-ii/

题目：

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.

You will have direct access to the node but not to the root of the tree. Each node will have a reference to its parent node.

Example 1:

Input:
root = {"$id":"1","left":{"$id":"2","left":null,"parent":{"$ref":"1"},"right":null,"val":1},"parent":null,"right":{"$id":"3","left":null,"parent":{"$ref":"1"},"right":null,"val":3},"val":2}
p = 1
Output: 2
Explanation: 1's in-order successor node is 2. Note that both p and the return value is of Node type.

Example 2:

Input:
root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":1},"parent":{"$ref":"2"},"right":null,"val":2},"parent":{"$ref":"1"},"right":{"$id":"5","left":null,"parent":{"$ref":"2"},"right":null,"val":4},"val":3},"parent":null,"right":{"$id":"6","left":null,"parent":{"$ref":"1"},"right":null,"val":6},"val":5}
p = 6
Output: null
Explanation: There is no in-order successor of the current node, so the answer is null.

Example 3:

Input:
root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":2},"parent":{"$ref":"2"},"right":{"$id":"5","left":null,"parent":{"$ref":"3"},"right":null,"val":4},"val":3},"parent":{"$ref":"1"},"right":{"$id":"6","left":null,"parent":{"$ref":"2"},"right":{"$id":"7","left":{"$id":"8","left":null,"parent":{"$ref":"7"},"right":null,"val":9},"parent":{"$ref":"6"},"right":null,"val":13},"val":7},"val":6},"parent":null,"right":{"$id":"9","left":{"$id":"10","left":null,"parent":{"$ref":"9"},"right":null,"val":17},"parent":{"$ref":"1"},"right":{"$id":"11","left":null,"parent":{"$ref":"9"},"right":null,"val":20},"val":18},"val":15}
p = 15
Output: 17

Example 4:

Input:
root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":2},"parent":{"$ref":"2"},"right":{"$id":"5","left":null,"parent":{"$ref":"3"},"right":null,"val":4},"val":3},"parent":{"$ref":"1"},"right":{"$id":"6","left":null,"parent":{"$ref":"2"},"right":{"$id":"7","left":{"$id":"8","left":null,"parent":{"$ref":"7"},"right":null,"val":9},"parent":{"$ref":"6"},"right":null,"val":13},"val":7},"val":6},"parent":null,"right":{"$id":"9","left":{"$id":"10","left":null,"parent":{"$ref":"9"},"right":null,"val":17},"parent":{"$ref":"1"},"right":{"$id":"11","left":null,"parent":{"$ref":"9"},"right":null,"val":20},"val":18},"val":15}
p = 13
Output: 15

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.
3. Remember that we are using the Node type instead of TreeNode type so their string representation are different.

Follow up:

Could you solve it without looking up any of the node's values?

题解：

Successor could exist in 2 possible positions.

If x has right child, successor must be below its right child, x.right, then keep going down left.

Otherwise, successor could be x going up untill hitting first ancestor through left edge. There may be case that it keeps going up through right edge, then there is no successor.

Time Complexity: O(h). h is the height of tree.

Space: O(1).

AC Java:

 /*
 // Definition for a Node.
 class Node {
     public int val;
     public Node left;
     public Node right;
     public Node parent;
 };
 */
 class Solution {
     public Node inorderSuccessor(Node x) {
         if(x == null){
             return x;
         }
 
         if(x.right != null){
             Node suc = x.right;
             while(suc.left != null){
                 suc = suc.left;
             }
 
             return suc;
         }
 
         while(x.parent != null && x.parent.right == x){
             x = x.parent;
         }
 
         return x.parent;
     }
 }