*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