数组实现二叉树

来源：互联网发布：2017软件开发工资编辑：程序博客网时间：2024/06/11 19:08

下面我们构造一个简单的二叉树，主要看清构造二叉树的思路；
解释一下，这里我们将二叉树的左节点和右边节点分别用0和1表示，同时将不存在的节点的值用0来表示.
binaryTree.h

class BinaryTree{public:    BinaryTree(int nSize, int *root);    bool addTreeNode(int pos, int* parent, int direct, int *node);    bool delTreeNode(int i);    void traverseTree();    ~BinaryTree();    friend bool cimi(int x);private:    int *m_Ptree;    int size_tree;};bool cimi(int x){    return (x&(x - 1)) ? true : false;}

createBinaryTree.cpp

#include <iostream>#include <cstdlib>#include <cmath>#include "binaryTree.h"using namespace std;BinaryTree::BinaryTree(int size, int *root){    m_Ptree = new int[size];    for (int i = 0; i < size; i++){        m_Ptree[i] = 0;    }    m_Ptree[0] = *root;    size_tree = size;}bool BinaryTree::delTreeNode(int i){    if (i < 0 || i >= size_tree){        return false;    }    if (m_Ptree[i] == 0){        return false;    }    m_Ptree[i] = 0;}bool BinaryTree::addTreeNode(int pos, int *parent,int direction,int *node){    if (direction == 0){        if (pos * 2 + 1 < 0 || pos * 2 + 1 >=size_tree){            cout << "左边超出" << endl;            return false;        }        if (m_Ptree[pos * 2 + 1] != 0){            cout << "左孩子已经有值" << endl;            return false;        }        cout << "第" << pos * 2 + 1 << "个子节点" << endl;        m_Ptree[pos * 2 + 1] = *node;    }    else{        if (pos * 2 + 2 < 0 || pos * 2 + 2 >= size_tree){            return false;        }        if (m_Ptree[pos * 2 + 2] != 0){            return false;        }        cout << "第" << pos * 2 + 2 << "个子节点" << endl;        m_Ptree[pos * 2 + 2] = *node;    }}void BinaryTree::traverseTree(){    for (int i = 0; i < size_tree; i++){        cout << m_Ptree[i] << " ";    }    cout << endl;}BinaryTree::~BinaryTree(){    delete[] m_Ptree;    m_Ptree = NULL;}int main(){    int root = 10;    BinaryTree bTree(7, &root);    int node1 = 7;    int node2 = 8;    bTree.addTreeNode(0, &root, 0, &node1);    bTree.addTreeNode(0, &root, 1, &node2);    int node3 = 6;    int node4 = 10;    bTree.addTreeNode(1, &node1, 0, &node3);    bTree.addTreeNode(1, &node1, 1, &node4);    int node5 = 100;    int node6 = 70;    bTree.addTreeNode(2, &node2, 0, &node5);    bTree.addTreeNode(2, &node2, 1, &node6);    bTree.traverseTree();    bTree.delTreeNode(6);    bTree.traverseTree();}

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