• 二叉树和链表复习


    今天根据斯坦福大学的在线课程复习链表和二叉树

    上面对于二叉树的递归定义非常简洁:

     a binary tree is either empty (represented by a null pointer), or is made of a single node, where the left and right pointers (recursive definition ahead) each point to a binary tree

    二叉树的一个分支:二叉搜索树(binary search tree,BST),是所有左节点的值 <= 中间节点 ,右节点的值 > 中间节点。

    这就是一棵二叉搜索树

    BST的优点是便于插入和查找,在一棵数量为N的BST中插入和查找的算法复杂度是O(lgN),类似二分查找。

    下面使用Java描述的二叉树,包括插入和查找的基本方法。另外说一点,面向对象的好处就是对外屏蔽不需要显示的细节,封装。

    // BinaryTree.java 
    public class BinaryTree { 
      // Root node pointer. Will be null for an empty tree. 
      private Node root; 
     
    
      /* 
       --Node-- 
       The binary tree is built using this nested node class. 
       Each node stores one data element, and has left and right 
       sub-tree pointer which may be null. 
       The node is a "dumb" nested class -- we just use it for 
       storage; it does not have any methods. 
      */ 
      private static class Node { 
        Node left; 
        Node right; 
        int data;
    
        Node(int newData) { 
          left = null; 
          right = null; 
          data = newData; 
        } 
      }
    
      /** 
       Creates an empty binary tree -- a null root pointer. 
      */ 
      public void BinaryTree() { 
        root = null; 
      } 
     
    
      /** 
       Returns true if the given target is in the binary tree. 
       Uses a recursive helper. 
      */ 
      public boolean lookup(int data) { 
        return(lookup(root, data)); 
      } 
     
    
      /** 
       Recursive lookup  -- given a node, recur 
       down searching for the given data. 
      */ 
      private boolean lookup(Node node, int data) { 
        if (node==null) { 
          return(false); 
        }
    
        if (data==node.data) { 
          return(true); 
        } 
        else if (data<node.data) { 
          return(lookup(node.left, data)); 
        } 
        else { 
          return(lookup(node.right, data)); 
        } 
      } 
     
    
      /** 
       Inserts the given data into the binary tree. 
       Uses a recursive helper. 
      */ 
      public void insert(int data) { 
        root = insert(root, data); 
      } 
     
    
      /** 
       Recursive insert -- given a node pointer, recur down and 
       insert the given data into the tree. Returns the new 
       node pointer (the standard way to communicate 
       a changed pointer back to the caller). 
      */ 
      private Node insert(Node node, int data) { 
        if (node==null) { 
          node = new Node(data); 
        } 
        else { 
          if (data <= node.data) { 
            node.left = insert(node.left, data); 
          } 
          else { 
            node.right = insert(node.right, data); 
          } 
        }
    
        return(node); // in any case, return the new pointer to the caller 
      } 

    另外还有获得树的高度以及最小值的方法

    //获得树的高度

    /**
    Returns the max root-to-leaf depth of the tree. Uses a recursive helper that recurs down to find the max depth. */ public int maxDepth() { return(maxDepth(root)); } private int maxDepth(Node node) { if (node==null) { return(0); } else { int lDepth = maxDepth(node.left); int rDepth = maxDepth(node.right); // use the larger + 1 return(Math.max(lDepth, rDepth) + 1); } } //获得最小值 /** Returns the min value in a non-empty binary search tree. Uses a helper method that iterates to the left to find the min value. */ public int minValue() { return( minValue(root) ); } /** Finds the min value in a non-empty binary search tree. */ private int minValue(Node node) { Node current = node; while (current.left != null) { current = current.left; } return(current.data); }

    关于链表

    记得大学还没毕业的时候有一次面试,面试官问我把一个链表倒置需要几个指针?当时没答出来,印象特别深刻,所以这次就把这个知识点复习下。

    使用C代码写的,但思路是一样的

    /*
     Iterative list reverse.
     Iterate through the list left-right.
    
     Move/insert each node to the front of the result list --
     like a Push of the node.
    */
    static void Reverse(struct node**headRef){
            struct node*result=NULL;
            struct node*current=*headRef;
            struct node*next;
            while(current!=NULL){
                next=current->next; // tricky: note the next node
                current->next=result; // move the node onto the result
                result=current;
                current=next;
            }
            
            *headRef=result;
    }

    需要三个指针,一个指向需要倒置的链表,一个保存next节点,一个指向倒置后返回的链表。

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  • 原文地址:https://www.cnblogs.com/andrew-chen/p/5481323.html
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