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c++先序二叉樹的構(gòu)建詳解

更新時間：2019年04月22日 08:41:26 投稿：laozhang

在本篇文章里小編給大家分享了關(guān)于c++先序二叉樹的構(gòu)建的相關(guān)知識點(diǎn)，需要的朋友們跟著學(xué)習(xí)下。

二叉樹首先要解決構(gòu)建問題，才能考慮后續(xù)的遍歷，這里貼出通過先序構(gòu)建二叉樹，同時包含四種二叉樹的遍歷方法（先序，中序，后序，逐層）

第一、定義BinaryTreeNode 類

#include <iostream>

#include <string>

#include <queue>

using namespace std;

 

template<typename T >class BinaryTree;

template <typename T> class BinaryTreeNode {

public:

  friend class BinaryTree<T>;

  BinaryTreeNode() {

    data = NULL;

    lChild = rChild = NULL;

  }

  BinaryTreeNode(T newdata) {

    this->data = newdata;

    lChild = rChild = NULL;

  }

  T getData() {

    return data;

  }

  BinaryTreeNode<T> * getLeftNode() {

    return lChild;

  }

  BinaryTreeNode<T> * getRightNode() {

    return rChild;

  }

  T data;

  BinaryTreeNode<T>* lChild;

  BinaryTreeNode<T>* rChild;

private:

 

};

View Code

第二、定義BinaryTree 類

template <typename T> class BinaryTree {

public:

  BinaryTreeNode<T> *root;

  char* p;

  BinaryTree() { root = NULL; }

  BinaryTree(T data) {

    root = new BinaryTreeNode<T>(data);

    root->lChild = NULL;

    root->rChild = NULL;

  }

  ~BinaryTree() {

    delete root;

  }

 

  //構(gòu)建二叉樹并返回

  BinaryTreeNode<T>* CreateTree() {

    BinaryTreeNode<int>* bt = NULL;

    char t;

    cin >> t;

    if (t == '#')

    {

      return NULL;

    }

    else {

      int num = t - '0';

      bt = new BinaryTreeNode<T>(num);

      bt->lChild = CreateTree();

      bt->rChild = CreateTree();

    }

    return bt;

  }

 

  //先序構(gòu)建二叉樹

  BinaryTreeNode<T>* PreCreateTree() {

    BinaryTreeNode<int>* bt = NULL;

    if (this->root == NULL)

    {

      cout << "請輸入根節(jié)點(diǎn)（#代表空樹）:";

    }

    else {

      cout << "請輸入節(jié)點(diǎn)（#代表空樹）:";

    }

    char t;

    cin >> t;

    if (t == '#')

    {

      return NULL;

    }

    else {

      int num = t - '0';

      bt = new BinaryTreeNode<T>(num);

      if (this->root == NULL)

      {

        this->root = bt;

      }

      cout << bt->data << "的左孩子";

      bt->lChild = PreCreateTree();

 

      cout << bt->data << "的右邊孩子";

      bt->rChild = PreCreateTree();

    }

    return bt;

  }  

 

  void preOderTraversal(BinaryTreeNode<T> *bt); //先序遍歷

  void inOrderTraversal(BinaryTreeNode<T> *bt); //中序遍歷

  void postOrderTraversal(BinaryTreeNode<T> *bt);//后序遍歷

  void levelTraversal(BinaryTreeNode<T> *bt);  //逐層遍歷

 

private:

 

};

 

template <typename T>

void BinaryTree<T>::preOderTraversal(BinaryTreeNode<T> *bt) {

  if (bt)

  {

    cout << bt->data;

    BinaryTree<T>::preOderTraversal(bt->getLeftNode());

    BinaryTree<T>::preOderTraversal(bt->getRightNode());

  }

}

 

template <typename T>

void BinaryTree<T>::inOrderTraversal(BinaryTreeNode<T> *bt) {

  if (bt)

  {

    BinaryTree<T>::inOrderTraversal(bt->getLeftNode());

    cout << bt->data;

    BinaryTree<T>::inOrderTraversal(bt->getRightNode());

  }

}

 

template <typename T>

void BinaryTree<T>::postOrderTraversal(BinaryTreeNode<T> *bt) {

  if (bt)

  {

    BinaryTree<T>::postOrderTraversal(bt->getLeftNode());

    BinaryTree<T>::postOrderTraversal(bt->getRightNode());

    cout << bt->data;

  }

}

 

template <typename T>

void BinaryTree<T>::levelTraversal(BinaryTreeNode<T> *bt) {

 

  queue<BinaryTreeNode<T>*> que;

  que.push(bt);

  while (!que.empty())

  {

    BinaryTreeNode<T>* proot = que.front();

    que.pop();

    cout << proot->data;

 

    if (proot->lChild != NULL)

    {

      que.push(proot->lChild);//左孩子入隊(duì)

    }

    if (proot->rChild != NULL)

    {

      que.push(proot->rChild);//右孩子入隊(duì)

    }

  }

}

View Code

第三、主程序運(yùn)行

#include "pch.h"

#include <iostream>

#include "BinaryTree.h"

 

int main()

{

  //場景測試2

  BinaryTree<int> btree;

  btree.PreCreateTree();//先序構(gòu)建二叉樹

  cout << "先序遍歷:";

  btree.preOderTraversal(btree.root); cout << endl;//先序遍歷  

  cout << "中序遍歷:";

  btree.inOrderTraversal(btree.root); cout << endl;//中序遍歷

  cout << "后序遍歷:";

  btree.postOrderTraversal(btree.root); cout << endl;//后序遍歷

  cout << "逐層序遍歷:";

  btree.levelTraversal(btree.root);

 

}

View Code

最終測試運(yùn)行截圖