修改二叉树以仅使用右指针获得先序遍历

2021年5月3日17:10:08 发表评论 1,166 次浏览

本文概述

给定二叉树。进行修改, 以便在修改后, 仅使用正确的指针就可以对其进行先序遍历。在修改期间, 你可以使用右指针和左指针。

例子：

Input :    10
          /  \
        8      2
      / \    
    3     5  
Output :    10
              \
               8
                \ 
                 3
                  \
                   5
                    \
                     2
Explanation : The preorder traversal
of given binary tree is 10 8 3 5 2.

推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。

方法1(递归)

需要使根的右指针指向左子树。

如果该节点刚好有左子节点, 则只需将子节点向右移动即可完成该节点的处理。

如果也有合适的孩子, 那么应该

原始左子树最右端的右子级。

代码中使用的上述函数处理一个节点, 然后返回转换后的子树的最右边的节点。

C ++

//C code to modify binary tree for
//traversal using only right pointer
#include <bits/stdc++.h>
  
using namespace std;
  
//A binary tree node has data, //left child and right child
struct Node {
     int data;
     struct Node* left;
     struct Node* right;
};
  
//function that allocates a new node
//with the given data and NULL left
//and right pointers.
struct Node* newNode( int data)
{
     struct Node* node = new struct Node;
     node->data = data;
     node->left = NULL;
     node->right = NULL;
     return (node);
}
  
//Function to modify tree
struct Node* modifytree( struct Node* root)
{
     struct Node* right = root->right;
     struct Node* rightMost = root;
  
     //if the left tree exists
     if (root->left) {
  
         //get the right-most of the
         //original left subtree
         rightMost = modifytree(root->left);
  
         //set root right to left subtree
         root->right = root->left;
         root->left = NULL;
     }
  
     //if the right subtree does
     //not exists we are done!
     if (!right) 
         return rightMost;
  
     //set right pointer of right-most
     //of the original left subtree
     rightMost->right = right;
  
     //modify the rightsubtree
     rightMost = modifytree(right);
     return rightMost;
}
  
//printing using right pointer only
void printpre( struct Node* root)
{
     while (root != NULL) {
         cout <<root->data <<" " ;
         root = root->right;
     }
}
  
//Driver program to test above functions
int main()
{
     /* Constructed binary tree is
          10
         /\
        8   2
       /\  
      3   5     */
     struct Node* root = newNode(10);
     root->left = newNode(8);
     root->right = newNode(2);
     root->left->left = newNode(3);
     root->left->right = newNode(5);
  
     modifytree(root);
     printpre(root);
  
     return 0;
}

Java

//Java code to modify binary tree for 
//traversal using only right pointer 
class GFG
{
  
//A binary tree node has data, //left child and right child 
static class Node 
{ 
     int data; 
     Node left; 
     Node right; 
}; 
  
//function that allocates a new node 
//with the given data and null left 
//and right pointers. 
static Node newNode( int data) 
{ 
     Node node = new Node(); 
     node.data = data; 
     node.left = null ; 
     node.right = null ; 
     return (node); 
} 
  
//Function to modify tree 
static Node modifytree( Node root) 
{ 
     Node right = root.right; 
     Node rightMost = root; 
  
     //if the left tree exists 
     if (root.left != null ) 
     { 
  
         //get the right-most of the 
         //original left subtree 
         rightMost = modifytree(root.left); 
  
         //set root right to left subtree 
         root.right = root.left; 
         root.left = null ; 
     } 
  
     //if the right subtree does 
     //not exists we are done! 
     if (right == null ) 
         return rightMost; 
  
     //set right pointer of right-most 
     //of the original left subtree 
     rightMost.right = right; 
  
     //modify the rightsubtree 
     rightMost = modifytree(right); 
     return rightMost; 
} 
  
//printing using right pointer only 
static void printpre( Node root) 
{ 
     while (root != null ) 
     { 
         System.out.print( root.data + " " ); 
         root = root.right; 
     } 
} 
  
//Driver cde 
public static void main(String args[]) 
{ 
     /* Constructed binary tree is 
         10 
         /\ 
     8 2 
     /\ 
     3 5 */
     Node root = newNode( 10 ); 
     root.left = newNode( 8 ); 
     root.right = newNode( 2 ); 
     root.left.left = newNode( 3 ); 
     root.left.right = newNode( 5 ); 
  
     modifytree(root); 
     printpre(root); 
}
} 
  
//This code is contributed by Arnab Kundu

Python3

# Python code to modify binary tree for 
# traversal using only right pointer 
  
class newNode(): 
  
     def __init__( self , data): 
         self .data = data
         self .left = None
         self .right = None
          
          
# Function to modify tree 
def modifytree(root):
  
     right = root.right 
     rightMost = root 
  
     # if the left tree exists 
     if (root.left):
  
         # get the right-most of the 
         # original left subtree 
         rightMost = modifytree(root.left) 
  
         # set root right to left subtree 
         root.right = root.left 
         root.left = None
      
  
     # if the right subtree does 
     # not exists we are done! 
     if ( not right):
         return rightMost 
  
     # set right pointer of right-most 
     # of the original left subtree 
     rightMost.right = right 
  
     # modify the rightsubtree 
     rightMost = modifytree(right) 
     return rightMost 
  
  
# printing using right pointer only 
def printpre(root):
  
     while (root ! = None ):
         print (root.data, end = " " )
         root = root.right 
          
# Driver code
if __name__ = = '__main__' :
     """ Constructed binary tree is
     10 
         /\ 
     8 2 
     /\ 
     3 5     """
     root = newNode( 10 ) 
     root.left = newNode( 8 ) 
     root.right = newNode( 2 ) 
     root.left.left = newNode( 3 ) 
     root.left.right = newNode( 5 ) 
  
     modifytree(root) 
     printpre(root)
  
# This code is contributed by SHUBHAMSINGH10

C#

//C# code to modify binary tree for 
//traversal using only right pointer
using System;
      
class GFG
{
  
//A binary tree node has data, //left child and right child 
public class Node 
{ 
     public int data; 
     public Node left; 
     public Node right; 
}; 
  
//function that allocates a new node 
//with the given data and null left 
//and right pointers. 
static Node newNode( int data) 
{ 
     Node node = new Node(); 
     node.data = data; 
     node.left = null ; 
     node.right = null ; 
     return (node); 
} 
  
//Function to modify tree 
static Node modifytree( Node root) 
{ 
     Node right = root.right; 
     Node rightMost = root; 
  
     //if the left tree exists 
     if (root.left != null ) 
     { 
  
         //get the right-most of the 
         //original left subtree 
         rightMost = modifytree(root.left); 
  
         //set root right to left subtree 
         root.right = root.left; 
         root.left = null ; 
     } 
  
     //if the right subtree does 
     //not exists we are done! 
     if (right == null ) 
         return rightMost; 
  
     //set right pointer of right-most 
     //of the original left subtree 
     rightMost.right = right; 
  
     //modify the rightsubtree 
     rightMost = modifytree(right); 
     return rightMost; 
} 
  
//printing using right pointer only 
static void printpre( Node root) 
{ 
     while (root != null ) 
     { 
         Console.Write( root.data + " " ); 
         root = root.right; 
     } 
} 
  
//Driver cde 
public static void Main(String []args) 
{ 
     /* Constructed binary tree is 
         10 
         /\ 
     8 2 
     /\ 
     3 5 */
     Node root = newNode(10); 
     root.left = newNode(8); 
     root.right = newNode(2); 
     root.left.left = newNode(3); 
     root.left.right = newNode(5); 
  
     modifytree(root); 
     printpre(root); 
}
}
  
//This code is contributed by 29AjayKumar

输出如下：

10 8 3 5 2

方法2(迭代)

使用迭代的先序遍历可以很容易地做到这一点。看这里。

迭代式遍历

这个想法是要维护一个变量prev, 它保持了先序遍历的前一个节点。每次遇到新节点时, 该节点将其权限设置为上一个节点, 并且上一个节点等于当前节点。最后, 我们将得到一种链表, 其第一个元素是root, 然后是left child, 然后是right, 依此类推。

C ++

//C++ code to modify binary tree for
//traversal using only right pointer
#include <iostream>
#include <stack>
#include <stdio.h>
#include <stdlib.h>
  
using namespace std;
  
//A binary tree node has data, //left child and right child
struct Node {
     int data;
     struct Node* left;
     struct Node* right;
};
  
//Helper function that allocates a new
//node with the given data and NULL
//left and right  pointers.
struct Node* newNode( int data)
{
     struct Node* node = new struct Node;
     node->data = data;
     node->left = NULL;
     node->right = NULL;
     return (node);
}
  
//An iterative process to set the right
//pointer of Binary tree
void modifytree( struct Node* root)
{
     //Base Case
     if (root == NULL)
         return ;
  
     //Create an empty stack and push root to it
     stack<Node*> nodeStack;
     nodeStack.push(root);
  
     /* Pop all items one by one. 
         Do following for every popped item
        a) print it
        b) push its right child
        c) push its left child
     Note that right child is pushed first
     so that left is processed first */
     struct Node* pre = NULL;
     while (nodeStack.empty() == false ) {
  
         //Pop the top item from stack
         struct Node* node = nodeStack.top();
  
         nodeStack.pop();
  
         //Push right and left children of
         //the popped node to stack
         if (node->right)
             nodeStack.push(node->right);
         if (node->left)
             nodeStack.push(node->left);
  
         //check if some previous node exists
         if (pre != NULL) {
  
             //set the right pointer of
             //previous node to currrent
             pre->right = node;
         }
  
         //set previous node as current node
         pre = node;
     }
}
  
//printing using right pointer only
void printpre( struct Node* root)
{
     while (root != NULL) {
         cout <<root->data <<" " ;
         root = root->right;
     }
}
  
//Driver code
int main()
{
     /* Constructed binary tree is
             10
           /  \
         8      2
       / \    
     3     5  
   */
     struct Node* root = newNode(10);
     root->left = newNode(8);
     root->right = newNode(2);
     root->left->left = newNode(3);
     root->left->right = newNode(5);
  
     modifytree(root);
     printpre(root);
  
     return 0;
}

Java

//Java code to modify binary tree for 
//traversal using only right pointer 
import java.util.*;
class GfG {
  
//A binary tree node has data, //left child and right child 
static class Node { 
     int data; 
     Node left; 
     Node right; 
}
  
//Helper function that allocates a new 
//node with the given data and NULL 
//left and right pointers. 
static Node newNode( int data) 
{ 
     Node node = new Node(); 
     node.data = data; 
     node.left = null ; 
     node.right = null ; 
     return (node); 
} 
  
//An iterative process to set the right 
//pointer of Binary tree 
static void modifytree(Node root) 
{ 
     //Base Case 
     if (root == null ) 
         return ; 
  
     //Create an empty stack and push root to it 
     Stack<Node> nodeStack = new Stack<Node> (); 
     nodeStack.push(root); 
  
     /* Pop all items one by one. 
         Do following for every popped item 
     a) print it 
     b) push its right child 
     c) push its left child 
     Note that right child is pushed first 
     so that left is processed first */
     Node pre = null ; 
     while (nodeStack.isEmpty() == false ) { 
  
         //Pop the top item from stack 
         Node node = nodeStack.peek(); 
  
         nodeStack.pop(); 
  
         //Push right and left children of 
         //the popped node to stack 
         if (node.right != null ) 
             nodeStack.push(node.right); 
         if (node.left != null ) 
             nodeStack.push(node.left); 
  
         //check if some previous node exists 
         if (pre != null ) { 
  
             //set the right pointer of 
             //previous node to currrent 
             pre.right = node; 
         } 
  
         //set previous node as current node 
         pre = node; 
     } 
} 
  
//printing using right pointer only 
static void printpre(Node root) 
{ 
     while (root != null ) { 
         System.out.print(root.data + " " ); 
         root = root.right; 
     } 
} 
  
//Driver code 
public static void main(String[] args) 
{ 
     /* Constructed binary tree is 
             10 
         /\ 
         8     2 
     /\     
     3     5 
*/
     Node root = newNode( 10 ); 
     root.left = newNode( 8 ); 
     root.right = newNode( 2 ); 
     root.left.left = newNode( 3 ); 
     root.left.right = newNode( 5 ); 
  
     modifytree(root); 
     printpre(root); 
  
}
}

python

# Python code to modify binary tree for 
# traversal using only right pointer 
  
# A binary tree node has data, # left child and right child 
class newNode(): 
  
     def __init__( self , data): 
         self .data = data 
         self .left = None
         self .right = None
  
# An iterative process to set the right 
# pointer of Binary tree 
def modifytree( root):
      
     # Base Case 
     if (root = = None ):
         return
          
     # Create an empty stack and append root to it 
     nodeStack = []
     nodeStack.append(root) 
      
     ''' Pop all items one by one. 
     Do following for every popped item 
     a) prit 
     b) append its right child 
     c) append its left child 
     Note that right child is appended first 
     so that left is processed first '''
     pre = None
     while ( len (nodeStack)):
          
         # Pop the top item from stack 
         node = nodeStack[ - 1 ]
         nodeStack.pop() 
          
         # append right and left children of 
         # the popped node to stack 
         if (node.right):
             nodeStack.append(node.right)
         if (node.left):
             nodeStack.append(node.left) 
              
         # check if some previous node exists 
         if (pre ! = None ):
              
             # set the right pointer of 
             # previous node to currrent 
             pre.right = node 
              
         # set previous node as current node 
         pre = node 
          
# printing using right pointer only 
def printpre( root):
     while (root ! = None ):
         print (root.data, end = " " )
         root = root.right 
      
# Driver code 
  
''' Constructed binary tree is 
         10 
     /\ 
     8     2 
/\     
3     5 
'''
root = newNode( 10 ) 
root.left = newNode( 8 ) 
root.right = newNode( 2 ) 
root.left.left = newNode( 3 ) 
root.left.right = newNode( 5 ) 
  
modifytree(root) 
printpre(root) 
  
# This code is contributed by SHUBHAMSINGH10

C#

//C# code to modify binary tree for 
//traversal using only right pointer 
using System; 
using System.Collections;
  
class GfG 
{
  
//A binary tree node has data, //left child and right child 
public class Node 
{ 
     public int data; 
     public Node left; 
     public Node right; 
}
  
//Helper function that allocates a new 
//node with the given data and NULL 
//left and right pointers. 
static Node newNode( int data) 
{ 
     Node node = new Node(); 
     node.data = data; 
     node.left = null ; 
     node.right = null ; 
     return (node); 
} 
  
//An iterative process to set the right 
//pointer of Binary tree 
static void modifytree(Node root) 
{ 
     //Base Case 
     if (root == null ) 
         return ; 
  
     //Create an empty stack and Push root to it 
     Stack nodeStack = new Stack(); 
     nodeStack.Push(root); 
  
     /* Pop all items one by one. 
         Do following for every Popped item 
     a) print it 
     b) Push its right child 
     c) Push its left child 
     Note that right child is Pushed first 
     so that left is processed first */
     Node pre = null ; 
     while (nodeStack.Count !=0) 
     { 
  
         //Pop the top item from stack 
         Node node = (Node)nodeStack.Peek(); 
  
         nodeStack.Pop(); 
  
         //Push right and left children of 
         //the Popped node to stack 
         if (node.right != null ) 
             nodeStack.Push(node.right); 
         if (node.left != null ) 
             nodeStack.Push(node.left); 
  
         //check if some previous node exists 
         if (pre != null ) 
         { 
  
             //set the right pointer of 
             //previous node to currrent 
             pre.right = node; 
         } 
  
         //set previous node as current node 
         pre = node; 
     } 
} 
  
//printing using right pointer only 
static void printpre(Node root) 
{ 
     while (root != null ) 
     { 
         Console.Write(root.data + " " ); 
         root = root.right; 
     } 
} 
  
//Driver code 
public static void Main(String []args) 
{ 
     /* Constructed binary tree is 
             10 
         /\ 
         8     2 
     /\     
     3     5 
*/
     Node root = newNode(10); 
     root.left = newNode(8); 
     root.right = newNode(2); 
     root.left.left = newNode(3); 
     root.left.right = newNode(5); 
  
     modifytree(root); 
     printpre(root); 
}
} 
  
//This code is contributed by
//Arnab Kundu

输出如下：

10 8 3 5 2
木子山

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