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.

第一种方法可以是中序遍历二叉搜索树，记录prevnode，如果等于p，记录当前的node。代码如下：

/** * 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) {        Stack<TreeNode> stack = new Stack<TreeNode>();        TreeNode res = null;        TreeNode prev = null;        while (root != null || !stack.isEmpty()) {            while (root != null) {                stack.push(root);                root = root.left;            }            root = stack.pop();            if (prev != null && prev == p) {                res = root;                break;            }            prev = root;            root = root.right;        }        return res;    }}

第二种方法根据二叉搜索树的性质解题。代码如下：

Successor

public TreeNode successor(TreeNode root, TreeNode p) {  if (root == null)    return null;  if (root.val <= p.val) {    return successor(root.right, p);  } else {    TreeNode left = successor(root.left, p);    return (left != null) ? left : root;  }}

Predecessor

public TreeNode predecessor(TreeNode root, TreeNode p) {  if (root == null)    return null;  if (root.val >= p.val) {    return predecessor(root.left, p);  } else {    TreeNode right = predecessor(root.right, p);    return (right != null) ? right : root;  }}