285. Inorder Successor in Binary Search Tree

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

 1 public class Solution {
 2 
 3     public TreeNode inorderSuccessor(TreeNode root, TreeNode n) {
 4         if (root == null) return null;
 5 
 6         if (n.right != null) {
 7             return min(n.right);
 8         }
 9 
10         TreeNode succ = null;
11         while (root != null) {
12             if (n.val < root.val) { //左子树的successor都是当前树的root。
13                 succ = root;
14                 root = root.left;
15             } else if (n.val > root.val)
16                 root = root.right;
17             else
18                 break;
19         }
20         return succ;
21     }
22 
23     public TreeNode min(TreeNode n) {
24         if (n.left != null) {
25             return min(n.left);
26         }
27         return n;
28     }
29 }
30 
31 class TreeNode {
32     TreeNode left;
33     TreeNode right;
34     int val;
35 
36     public TreeNode(int i) {
37         val = i;
38     }
39 }

 Another solution

 1 class Solution {
 2     public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
 3         return helper(root, p, null);
 4     }
 5     
 6     private TreeNode helper(TreeNode root, TreeNode p, TreeNode parent) {
 7         if (root == null) {
 8             return parent;
 9         }
10         
11         if (root.val > p.val) {
12             return helper(root.left, p, root);
13         } else {
14             return helper(root.right, p, parent);
15         }
16     }
17 }

 

posted @ 2016-08-07 00:05  北叶青藤  阅读(200)  评论(0)  编辑  收藏  举报