Inorder Successor in BST
題意：
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.
解題思路：
是要找bst中某個節點的中序後繼者，如果是往左走，則當下的root有可能是該節點的後繼者，需紀錄下來，若是往右走，則後繼者不變，程式碼如下：
/**
* 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) {
return root;
}
TreeNode succ = null;
while (root != null) {
if (p.val < root.val) {
succ = root;
root = root.left;
} else {
root = root.right;
}
}
return succ;
}
}