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  • 二叉树的遍历

    #include<iostream>
    #include
    <assert.h>
    #include
    <stack>
    #include
    <queue>
    using namespace std;
    struct Node
    {
    int v;
    Node
    *leftChild,*rightChild;
    Node():leftChild(NULL),rightChild(NULL){}
    Node(
    int vv):leftChild(NULL),rightChild(NULL)
    {
    v
    =vv;
    }
    };

    void print(int v)
    {
    cout
    <<v<<" ";
    }
    void PreOrderTraverse(Node *n, void (* visit)(int))
    {
    assert(n
    !=NULL&&visit!=NULL);
    (
    *visit)(n->v);
    if(n->leftChild!=NULL) PreOrderTraverse(n->leftChild,visit);
    if(n->rightChild!=NULL) PreOrderTraverse(n->rightChild,visit);
    }

    void InOrderTraverse(Node *n, void (* visit)(int))
    {
    assert(n
    !=NULL&&visit!=NULL);
    if(n->leftChild!=NULL) InOrderTraverse(n->leftChild,visit);
    (
    *visit)(n->v);
    if(n->rightChild!=NULL) InOrderTraverse(n->rightChild,visit);
    }

    void PostOrderTraverse(Node *n, void (* visit)(int))
    {
    assert(n
    !=NULL&&visit!=NULL);
    if(n->leftChild!=NULL) PostOrderTraverse(n->leftChild,visit);
    if(n->rightChild!=NULL) PostOrderTraverse(n->rightChild,visit);
    (
    *visit)(n->v);
    }
    //非递归版本,将递归改成非递归一般都要利用一个栈
    //每次访问一个结点后,在向左子树遍历下去之前,利用这个栈记录该结点的右子女(如果有的话)结点的地址,
    //以便在左子树退回时可以直接从栈顶取得右子树的根结点,继续右子树的遍历
    void PreOrder(Node *n, void (* visit)(int))
    {
    stack
    <Node*> sta;
    sta.push(n);
    while(!sta.empty())
    {
    Node
    * t=sta.top();
    sta.pop();
    assert(t
    !=NULL);
    (
    *visit)(t->v);
    if(t->rightChild!=NULL) sta.push(t->rightChild);
    if(t->leftChild!=NULL) sta.push(t->leftChild);
    }
    }

    //非递归中序遍历
    void InOrder(Node * n , void (* visit) (int))
    {
    stack
    <Node *> sta;
    sta.push(n);
    Node
    * p= n;
    while(!sta.empty()&&p!=NULL)
    {
    p
    =sta.top();
    while(p!=NULL&&!sta.empty())
    {
    sta.push(p
    ->leftChild);
    p
    =p->leftChild;
    }
    sta.pop();
    //弹出空指针
    if(!sta.empty())
    {
    p
    =sta.top();
    sta.pop();
    (
    *visit)(p->v);
    sta.push(p
    ->rightChild);
    }
    }
    }


    //非递归后续遍历

    struct StkNode
    {
    Node
    * ptr;
    bool tag;//false=left and true=right
    StkNode():ptr(NULL),tag(false)
    {}
    };
    void PostOrder(Node * n ,void (*visit) (int))
    {
    stack
    <StkNode> sta;
    StkNode w;
    Node
    * p = n;
    do {
    while(p!=NULL)
    {
    w.ptr
    =p;
    w.tag
    =false;
    sta.push(w);
    p
    =p->leftChild;
    }
    bool flag=true;
    while(flag&&!sta.empty())
    {
    w
    =sta.top();
    sta.pop();
    p
    =w.ptr;
    if(!w.tag)//left,如果从左子树返回,则开始遍历右子树
    {
    w.tag
    =true;//标记右子树
    sta.push(w);
    flag
    =false;
    p
    =p->rightChild;
    }
    else
    {
    (
    *visit)(p->v);
    }
    }
    }
    while(!sta.empty());
    }

    //层序遍历,利用队列
    void LevelOrderTraverse(Node * n , void (* visit )(int))
    {
    assert(n
    !=NULL&&visit!=NULL);
    queue
    <Node * > que;
    que.push(n);
    while(!que.empty())
    {
    Node
    * t=que.front();
    (
    *visit)(t->v);
    que.pop();
    if(t->leftChild!=NULL) que.push(t->leftChild);
    if(t->rightChild!=NULL) que.push(t->rightChild);
    }
    }

    int main()
    {
    Node
    * head= new Node(0);
    Node
    * node1= new Node(1);
    Node
    * node2= new Node(2);
    Node
    * node3= new Node(3);
    Node
    * node4= new Node(4);
    Node
    * node5= new Node(5);
    Node
    * node6= new Node(6);


    head
    ->leftChild=node1;
    head
    ->rightChild=node2;
    node1
    ->leftChild=node3;
    node1
    ->rightChild=node4;
    node2
    ->rightChild=node5;
    node4
    ->leftChild=node6;


    /* LevelOrderTraverse(head,print);
    cout<<endl;
    PreOrderTraverse(head,print);
    cout<<endl;
    */
    InOrder(head,print);
    cout
    <<endl;
    InOrderTraverse(head,print);
    cout
    <<endl;

    PostOrder(head,print);
    cout
    <<endl;
    PostOrderTraverse(head,print);
    cout
    <<endl;
    return 0;
    }
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  • 原文地址:https://www.cnblogs.com/coser/p/2011107.html
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