二叉树的前、中、后序遍历的实现

简介: 二叉树的前、中、后序遍历的实现


<你想看的我这里都有😎 >

二叉树的遍历

概念:二叉树遍历是按照某种特定的规则,依次对二叉树中的节点进行相应的操作,并且每个节点只操作一次遍历是二叉树上最重要的运算之一,也是二叉树上进行其它运算的基础。

二叉树的递归定义:二叉树是一棵空树,或者是一棵由一个根节点和两棵互不相交的,分别称作根的左子树和右子树组成的非空树,左子树和右子树又同样都是二叉树。

包含内容:

  1. 前序遍历——访问根结点的操作发生在遍历其左右子树之前,即根->左->右
  2. 中序遍历——访问根结点的操作发生在遍历其左右子树之中,左->根->右
  3. 后序遍历——访问根结点的操作发生在遍历其左右子树之后,左->右->根

为了方便学习,我们将空的位置仍然以NULL表示

模拟实现二叉树(链式二叉树)

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
typedef int BTDataType;
typedef struct BinaryTreeNode
{
  BTDataType data;
  struct BinaryTreeNode* left;
  struct BinaryTreeNode* right;
}TreeNode;
TreeNode* BuyTreeNode(int x)
{
  TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
  assert(node);
  node->data = x;
  node->left = NULL;
  node->right = NULL;
  return node;
}
TreeNode* CreatTree()
{
  TreeNode* node1 = BuyTreeNode(1);
  TreeNode* node2 = BuyTreeNode(2);
  TreeNode* node3 = BuyTreeNode(3);
  TreeNode* node4 = BuyTreeNode(4);
  TreeNode* node5 = BuyTreeNode(5);
  TreeNode* node6 = BuyTreeNode(6);
  node1->left = node2;
  node1->right = node4;
  node2->left = node3;
  //node2->right = NULL;
  //node3->left = NULL;
  //node3->right = NULL;
  node4->left = node5;
  node4->right = node6;
  //node5->left = NULL;
  //node5->right = NULL;
  //node6->left = NULL;
  //node6->right= NULL;
  return node1;
}
int main()
{
  TreeNode* root = CreatTree();
  return 0;
}

前序遍历

前序遍历的实现

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
typedef int BTDataType;
typedef struct BinaryTreeNode
{
  BTDataType data;
  struct BinaryTreeNode* left;
  struct BinaryTreeNode* right;
}TreeNode;
TreeNode* BuyTreeNode(int x)
{
  TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
  assert(node);
  node->data = x;
  node->left = NULL;
  node->right = NULL;
  return node;
}
TreeNode* CreatTree()
{
  TreeNode* node1 = BuyTreeNode(1);
  TreeNode* node2 = BuyTreeNode(2);
  TreeNode* node3 = BuyTreeNode(3);
  TreeNode* node4 = BuyTreeNode(4);
  TreeNode* node5 = BuyTreeNode(5);
  TreeNode* node6 = BuyTreeNode(6);
  node1->left = node2;
  node1->right = node4;
  node2->left = node3;
  //node2->right = NULL;
  //node3->left = NULL;
  //node3->right = NULL;
  node4->left = node5;
  node4->right = node6;
  //node5->left = NULL;
  //node5->right = NULL;
  //node6->left = NULL;
  //node6->right= NULL;
  return node1;
}
void PrevOrder(TreeNode* root)
{
  if(root == NULL)
  {
    printf("N ");
    return;
  }
  printf("%d ", root->data);
  PrevOrder(root->left);
  PrevOrder(root->right);
}
int main()
{
  TreeNode* root = CreatTree();
  PrevOrder(root);
  return 0;
}

实现步骤:

1、在完成一个二叉树的基础上,我们实现了前序遍历的理论变现,即根->左->右

2、先令前序遍历函数接收总根节点的值(单链表的头)

3、进入函数后判断此时的根节点(1)是否为空,若为空则打印N表示空,若不为空则打印对应的值(1)

4、此后递归的读取当前结点(1)的左子树的根节点(3),若该结点不为空则打印对应的值(3),然后继续递归读取其左子树的根节点(NULL),此时结点为空打印N后返回上一次的位置(3),然后执行当前结点(3)的右递归,当前结点的右子树的根节点(NULL)为空打印N后结束,返回上一次的位置(2)(到这里整个左子树最底层的一个PrevOrder已经结束了),然后执行当前结点(2)的右递归,当前结点的右子树的根节点(NULL)为空打印N后结束,返回上一次位置(1)(到这里整个左子树的倒数第二个PrevOrder已经结束了)

5、接下俩就应该去读取(1)的右子树的根节点(4)了(因为到这里第一个PrevOrder的左递归才算完全结束,现在开始右递归)

6、(4)不为空打印对应的值(4),然后继续递归读取其左子树的根节点(5),不为空打印对应的值(5),然后继续递归读取其左子树的根节点(NULL)为空打印N后返回上一次的位置(5),然后继续递归读取其右子树的根节点(NULL)为空打印N后返回上一次的位置(4)(到这里整个右子最底层的一个PrevOrder已经结束了)然后继续递归读取其右子树的根节点(6),不为空打印对应的值(6),然后继续递归读取其左子树的根节点(NULL)为空打印N后返回上一次的位置(6),然后继续递归读取其右子树的根节点(NULL)为空后打印N后返回上一次的位置(4)

7、至此,该二叉树的前序遍历完成

一共用了五次PrevOrder,左子树两次,右子树两次,主二叉树一次

二叉树的递归遍历图


中序遍历

中序遍历的实现

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
typedef int BTDataType;
typedef struct BinaryTreeNode
{
  BTDataType data;
  struct BinaryTreeNode* left;
  struct BinaryTreeNode* right;
}TreeNode;
TreeNode* BuyTreeNode(int x)
{
  TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
  assert(node);
  node->data = x;
  node->left = NULL;
  node->right = NULL;
  return node;
}
TreeNode* CreatTree()
{
  TreeNode* node1 = BuyTreeNode(1);
  TreeNode* node2 = BuyTreeNode(2);
  TreeNode* node3 = BuyTreeNode(3);
  TreeNode* node4 = BuyTreeNode(4);
  TreeNode* node5 = BuyTreeNode(5);
  TreeNode* node6 = BuyTreeNode(6);
  node1->left = node2;
  node1->right = node4;
  node2->left = node3;
  //node2->right = NULL;
  //node3->left = NULL;
  //node3->right = NULL;
  node4->left = node5;
  node4->right = node6;
  //node5->left = NULL;
  //node5->right = NULL;
  //node6->left = NULL;
  //node6->right= NULL;
  return node1;
}
void InOrder(TreeNode* root)
{
  if (root == NULL)
  {
    printf("N ");
    return;
  }
  InOrder(root->left);
  printf("%d ", root->data);
  InOrder(root->right);
}
int main()
{
  TreeNode* root = CreatTree();
  InOrder(root);
  return 0;
}

实现步骤:中序遍历只是将前序遍历的PrevOrder中的 (root->data)与 (root->left)交换了位置,具体实现内容建议自行尝试


后序遍历

后序遍历的实现

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
typedef int BTDataType;
typedef struct BinaryTreeNode
{
  BTDataType data;
  struct BinaryTreeNode* left;
  struct BinaryTreeNode* right;
}TreeNode;
TreeNode* BuyTreeNode(int x)
{
  TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
  assert(node);
  node->data = x;
  node->left = NULL;
  node->right = NULL;
  return node;
}
TreeNode* CreatTree()
{
  TreeNode* node1 = BuyTreeNode(1);
  TreeNode* node2 = BuyTreeNode(2);
  TreeNode* node3 = BuyTreeNode(3);
  TreeNode* node4 = BuyTreeNode(4);
  TreeNode* node5 = BuyTreeNode(5);
  TreeNode* node6 = BuyTreeNode(6);
  node1->left = node2;
  node1->right = node4;
  node2->left = node3;
  //node2->right = NULL;
  //node3->left = NULL;
  //node3->right = NULL;
  node4->left = node5;
  node4->right = node6;
  //node5->left = NULL;
  //node5->right = NULL;
  //node6->left = NULL;
  //node6->right= NULL;
  return node1;
}
void LaterOrder(TreeNode* root)
{
  if (root == NULL)
  {
    printf("N ");
    return;
  }
  LaterOrder(root->left);
  LaterOrder(root->right);
  printf("%d ", root->data);
}
int main()
{
  TreeNode* root = CreatTree();
  LaterOrder(root);
  printf("\n");
  return 0;
}

注意事项:

1、建议多画图自己尝试一遍

2、搞懂此时的root是谁以及root->data到底打印的是谁的值,是完成三次遍历的基础


整体代码

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
typedef int BTDataType;
typedef struct BinaryTreeNode
{
  BTDataType data;
  struct BinaryTreeNode* left;
  struct BinaryTreeNode* right;
}TreeNode;
TreeNode* BuyTreeNode(int x)
{
  TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
  assert(node);
  node->data = x;
  node->left = NULL;
  node->right = NULL;
  return node;
}
TreeNode* CreatTree()
{
  TreeNode* node1 = BuyTreeNode(1);
  TreeNode* node2 = BuyTreeNode(2);
  TreeNode* node3 = BuyTreeNode(3);
  TreeNode* node4 = BuyTreeNode(4);
  TreeNode* node5 = BuyTreeNode(5);
  TreeNode* node6 = BuyTreeNode(6);
  node1->left = node2;
  node1->right = node4;
  node2->left = node3;
  //node2->right = NULL;
  //node3->left = NULL;
  //node3->right = NULL;
  node4->left = node5;
  node4->right = node6;
  //node5->left = NULL;
  //node5->right = NULL;
  //node6->left = NULL;
  //node6->right= NULL;
  return node1;
}
void PrevOrder(TreeNode* root)
{
  if(root == NULL)
  {
    printf("N ");
    return;
  }
  printf("%d ", root->data);
  PrevOrder(root->left);
  PrevOrder(root->right);
}
void InOrder(TreeNode* root)
{
  if (root == NULL)
  {
    printf("N ");
    return;
  }
  InOrder(root->left);
  printf("%d ", root->data);
  InOrder(root->right);
}
void LaterOrder(TreeNode* root)
{
  if (root == NULL)
  {
    printf("N ");
    return;
  }
  LaterOrder(root->left);
  LaterOrder(root->right);
  printf("%d ", root->data);
}
int main()
{
  TreeNode* root = CreatTree();
  PrevOrder(root);
  printf("\n");
  InOrder(root);
  printf("\n");
  LaterOrder(root);
  printf("\n");
  return 0;
}

~over~

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