LeetCode 285. Inorder Successor in BST

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

题目：

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

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

题解：

successor could be自己的parent, or 右子树中最左的点.

For both cases, if p.val < root.val, update successor as root. Otherwise, root needs to be bigger, root = root.right.

Time Complexity: O(h).

Space: O(1).

AC Java:

/**
 * 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) {
        if(root == null || p == null){
            return null;
        }
        TreeNode successor = null;
        while(root != null){
            if(p.val < root.val){
                successor = root;
                root = root.left;
            }else{
                root = root.right;
            }
        }
        return successor;
    }
}

跟上Inorder Successor in BST II.

类似Binary Search Tree Iterator.