285. 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.

做出来了 但是需要遍历整棵树 

下面的基于二分查找 针对后继节点出现的情况进行分类

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;
}

也可以从二叉树的中序遍历是有序的来考虑 就是找所有大于p的最小节点