zoukankan      html  css  js  c++  java
  • Inorder Successor in BST 解答

    Question

    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.

    Solution -- Iterative

    Inorder result is an ascending array for BST. Thus, this problem can be transferred to find successor for a node in BST.

    successor = first element that is greater than target node.

    Consider two situations:

    1. Target node has right child

    2. Target node doesn't have right child

    Trace back, find first node whose left child is in the path from target node to root.

    For this situation:

      1. If node has parent pointer, we just use parent pointer.

      2. If node doesn't have parent pointer, we use stack.

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
    12         // Same question with finding successor in a BST
    13         // 1. If p has right child, find minimum child in right subtree
    14         // 2. If p doesn't have right child, find first ancestor whose left child is in the path from p to root.
    15         TreeNode result = null;
    16         if (p.right != null) {
    17             result = p.right;
    18             while (result.left != null) {
    19                 result = result.left;
    20             }
    21         } else {
    22             Deque<TreeNode> stack = new LinkedList<TreeNode>();
    23             TreeNode current = root;
    24             // push to stack
    25             while (current != p && current != null) {
    26                 stack.push(current);
    27                 if (p.val < current.val) {
    28                     current = current.left;
    29                 } else {
    30                     current = current.right;
    31                 }
    32             }
    33             // pop stack
    34             TreeNode prev = p;
    35             while (!stack.isEmpty()) {
    36                 TreeNode cur = stack.pop();
    37                 if (cur.left == prev) {
    38                     result = cur;
    39                     break;
    40                 }
    41                 prev = cur;
    42             }
    43         }
    44         return result;
    45     }
    46 }

    Solution 2 -- Recursive

    Find Successor & Predecessor in BST

  • 相关阅读:
    Spring5源码--Spring AOP使用接口方式实现 配置xml文件
    什么是 JavaConfig?
    Spring Boot 有哪些优点?
    什么是 Spring Boot?
    问:一台客户端有三百个客户与三百个客户端有三百个客户对服务器施压,有什么区别?
    如何实现参数级联查询
    如何开发主从报表
    如何在报表中实现算法的可挂接需求
    如何实现报表的批量打印需求
    如何实现参数和报表间的联动效果
  • 原文地址:https://www.cnblogs.com/ireneyanglan/p/4946830.html
Copyright © 2011-2022 走看看