LintCode : Inorder Successor in Binary Search Tree

Very interesting problem!

http://www.lintcode.com/zh-cn/problem/inorder-successor-in-binary-search-tree/

Description:

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

Example:

Given tree = [2,1] and node = 1:

  2
 /
1

return node 2.

Given tree = [2,1,3] and node = 2:

  2
 / \
1   3

return node 3.

Challenge:

O(h), where h is the height of the BST.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        // write your code here
        TreeNode successor = null;
        while (root != null && root.val != p.val) {
            if (root.val > p.val) {
                successor = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }
        if (root == null) {
            return null;
        }
        if (root.right == null) {
            return successor;
        }
        root = root.right;
        while (root.left != null) {
            root = root.left;
        }
        return root;
    }
}

三种情况：

1，p有右儿子，这个时候就找右儿子一直找左儿子，直到null，可以根据in order的算法来想

2，p没有右儿子，这个时候就要找p的祖先，这个时候存在两种情况，如果p对p的祖先而言都是右儿子，这个时候就没有in order successor了，因为p就是in order的最后一个元素，如果不是这样，那么就找最近的对p而言是p是左儿子的，所以在这个程序中初始化successor为null，然后只在root.val > p.val时才更新successor,这两点很重要！