*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.

解法一：俺自个儿的方法，居然跑了16ms。。。妈蛋！

public class Solution {
    public TreeNode inorderSuccessor(TreeNode root, TreeNode p)
    {
        ArrayList<TreeNode> res = inOrderTrav(root);
        for(int i=0;i<res.size()-1;i++)
        {
            if(p==res.get(i))
            return res.get(i+1);
        }

        return null;

    }

    public ArrayList<TreeNode> inOrderTrav(TreeNode root)
    {
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        ArrayList<TreeNode> res = new ArrayList<TreeNode>();
        if(root==null) return res;
        while(root!=null||!stack.isEmpty())
        {
            if(root!=null)
            {
                stack.push(root);
                root = root.left;
            }
            else
            {
                root = stack.pop();
                res.add(root);
                root = root.right;
            }
        }
        return res;
    }
}

解法二：

public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
    TreeNode succ = null;
    while (root != null) {
        if (p.val < root.val) {
            succ = root;
            root = root.left;
        }
        else
            root = root.right;
    }
    return succ;
}

reference：https://leetcode.com/discuss/61105/java-python-solution-o-h-time-and-o-1-space-iterative