Inorder Successor in BST

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.

Example 1:

Input: root = [2,1,3], p = 1

  2
 / \
1   3

Output: 2

Example 2:

Input: root = [5,3,6,2,4,null,null,1], p = 6

      5
     / \
    3   6
   / \
  2   4
 /   
1

Output: null

time O(logN) space o(1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
        if(root == null || p == null) return null;
        
        TreeNode successor = null;
        
        
        //找p，记录successor
        while(root != null && root.val != p.val){
            if(root.val > p.val){
                successor = root;
                root = root.left;
            }else{
                root = root.right;
            }
        }
        
        //判断有没有找到p
        if(root == null)
            return null;
        
        
        root = root.right;
        
       if(root == null) return successor;
        
        //找右子树的最小值
        while(root.left != null){
            root = root.left;
        }
        
        return root;
        
    }
}