Inorder Successor in Binary Search Tree(lintcode 448)

Description

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

If the given node has no in-order successor in the tree, return null.

Notice:
It's guaranteed p is one node in the given tree. (You can directly compare the memory address to find p)

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.

Interface

/**
 * 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
    }
}

Solution

/**
 * 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 != p) {
            if (root.val > p.val) {
                successor = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }

        if (root == null) {
            return successor;
        }

        if (root.right == null) {
            return successor;
        }

        root = root.right;
        while (root.left != null) {
            root = root.left;
        }

        return root;
    }
}