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.
pp.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:
null.- It's guaranteed that the values of the tree are unique.
NodeTreeNode
Follow up:
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:
1 /* 2 // Definition for a Node. 3 class Node { 4 public int val; 5 public Node left; 6 public Node right; 7 public Node parent; 8 }; 9 */ 10 class Solution { 11 public Node inorderSuccessor(Node x) { 12 if(x == null){ 13 return x; 14 } 15 16 if(x.right != null){ 17 Node suc = x.right; 18 while(suc.left != null){ 19 suc = suc.left; 20 } 21 22 return suc; 23 } 24 25 while(x.parent != null && x.parent.right == x){ 26 x = x.parent; 27 } 28 29 return x.parent; 30 } 31 }