[LintCode] 448 Inorder Successor in Binary Search Tree 解题报告

Description
Given a binary search tree (See Definition) and a node in it, find the in-order successor of that node in the BST.

If the given node has no in-order successor in the tree, return null.


Notice
It's guaranteed p is one node in the given tree. (You can directly compare the memory address to find p)



Example
Given tree = [2,1] and node = 1:

  2
 /
1
return node 2.

Given tree = [2,1,3] and node = 2:

  2
 / \
1   3
return node 3.



Challenge
O(h), where h is the height of the BST.



思路
中序遍历是左根右的顺序。
在某一点root，如果root.val > p.val，说明p在root的左子树。那么我们可以往左继续找p。
这时root就是候选的successor。
在某一点root，如果root.val < p.val，说明p在root的右子树。那么我们保持现在的候选successor。
总结：只要往左走，就更新successor的候选为当前点。




Code

/**
 * 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) {
        // write your code here
        
        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;
    }
}