zoukankan      html  css  js  c++  java
  • LeetCode 510. Inorder Successor in BST II

    原题链接在这里:https://leetcode.com/problems/inorder-successor-in-bst-ii/

    题目:

    Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

    The successor of a node p is the node with the smallest key greater than p.val.

    You will have direct access to the node but not to the root of the tree. Each node will have a reference to its parent node.

    Example 1:

    Input: 
    root = {"$id":"1","left":{"$id":"2","left":null,"parent":{"$ref":"1"},"right":null,"val":1},"parent":null,"right":{"$id":"3","left":null,"parent":{"$ref":"1"},"right":null,"val":3},"val":2}
    p = 1
    Output: 2
    Explanation: 1's in-order successor node is 2. Note that both p and the return value is of Node type.
    

    Example 2:

    Input: 
    root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":1},"parent":{"$ref":"2"},"right":null,"val":2},"parent":{"$ref":"1"},"right":{"$id":"5","left":null,"parent":{"$ref":"2"},"right":null,"val":4},"val":3},"parent":null,"right":{"$id":"6","left":null,"parent":{"$ref":"1"},"right":null,"val":6},"val":5}
    p = 6
    Output: null
    Explanation: There is no in-order successor of the current node, so the answer is null.
    

    Example 3:

    Input: 
    root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":2},"parent":{"$ref":"2"},"right":{"$id":"5","left":null,"parent":{"$ref":"3"},"right":null,"val":4},"val":3},"parent":{"$ref":"1"},"right":{"$id":"6","left":null,"parent":{"$ref":"2"},"right":{"$id":"7","left":{"$id":"8","left":null,"parent":{"$ref":"7"},"right":null,"val":9},"parent":{"$ref":"6"},"right":null,"val":13},"val":7},"val":6},"parent":null,"right":{"$id":"9","left":{"$id":"10","left":null,"parent":{"$ref":"9"},"right":null,"val":17},"parent":{"$ref":"1"},"right":{"$id":"11","left":null,"parent":{"$ref":"9"},"right":null,"val":20},"val":18},"val":15}
    p = 15
    Output: 17
    

    Example 4:

    Input: 
    root = {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":{"$id":"4","left":null,"parent":{"$ref":"3"},"right":null,"val":2},"parent":{"$ref":"2"},"right":{"$id":"5","left":null,"parent":{"$ref":"3"},"right":null,"val":4},"val":3},"parent":{"$ref":"1"},"right":{"$id":"6","left":null,"parent":{"$ref":"2"},"right":{"$id":"7","left":{"$id":"8","left":null,"parent":{"$ref":"7"},"right":null,"val":9},"parent":{"$ref":"6"},"right":null,"val":13},"val":7},"val":6},"parent":null,"right":{"$id":"9","left":{"$id":"10","left":null,"parent":{"$ref":"9"},"right":null,"val":17},"parent":{"$ref":"1"},"right":{"$id":"11","left":null,"parent":{"$ref":"9"},"right":null,"val":20},"val":18},"val":15}
    p = 13
    Output: 15
    

    Note:

    1. If the given node has no in-order successor in the tree, return null.
    2. It's guaranteed that the values of the tree are unique.
    3. Remember that we are using the Node type instead of TreeNode type so their string representation are different.

    Follow up:

    Could you solve it without looking up any of the node's values?

    题解:

    Successor could exist in 2 possible positions.

    If x has right child, successor must be below its right child, x.right, then keep going down left.

    Otherwise, successor could be x going up untill hitting first ancestor through left edge. There may be case that it keeps going up through right edge, then there is no successor.

    Time Complexity: O(h). h is the height of tree.

    Space: O(1).

    AC Java:

     1 /*
     2 // Definition for a Node.
     3 class Node {
     4     public int val;
     5     public Node left;
     6     public Node right;
     7     public Node parent;
     8 };
     9 */
    10 class Solution {
    11     public Node inorderSuccessor(Node x) {
    12         if(x == null){
    13             return x;
    14         }
    15         
    16         if(x.right != null){
    17             Node suc = x.right;
    18             while(suc.left != null){
    19                 suc = suc.left;
    20             }
    21             
    22             return suc;
    23         }
    24         
    25         while(x.parent != null && x.parent.right == x){
    26             x = x.parent;
    27         }
    28         
    29         return x.parent;
    30     }
    31 }
  • 相关阅读:
    util-tool工具 、时间格式化
    好用的 UI 调试技巧
    扩展jQuery的方法
    mybatis判断是否传递了条件
    mysql创建视图不允许子查询
    springMVC传递MAP给JS
    XPS15 安装ubuntu之后无法进入系统
    XPS15 9560 切换显卡之后,无法登陆的解决方式
    fedora安装chrome报错
    linux挂载硬盘
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11015316.html
Copyright © 2011-2022 走看看