二叉树复习

最新推荐文章于 2024-12-31 23:21:45 发布

原创最新推荐文章于 2024-12-31 23:21:45 发布 · 217 阅读

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#二叉树

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复习二叉树的前中后序遍历非递归，有完整代码，也有在leetcode上的函数部分，第一次用vector
另外，->和.是不一样的
比如，struct p p.a
struct *p p->a
关于bool isLeaf()const
这个是const成员函数，为了维持程序的健壮性
保证了函数在执行过程中不会修改数据成员，也不允许调用非const成员函数

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> st;
        vector<int>list;
        TreeNode*current=root;
        int count=0;
        int p[200];
        
        while(!st.empty()||current!=NULL)
        {
            while(current!=NULL)
            {
                st.push(current);
                list.push_back(current->val);
                //p[count++]=current->val;
                current=current->left;
            }
            if(!st.empty())
            {
                current=st.top();
                st.pop();
                current=current->right;
            }
        }
        return list;
        //return p;
    }
};

完整的三种遍历的代码，后序遍历比较麻烦，加isFirst进行判断

#include<stdio.h>
#include<iostream>
#include<stack>
using namespace std;
template<class T>
class BinaryTree;
template<class T>
class TreeNode
{
    friend class BinaryTree<T>;
public:
    T data;
    bool isFirst;
    TreeNode<T>*lchild;
    TreeNode<T>*rchild;
    TreeNode(T da,TreeNode<T>*l=NULL,TreeNode<T>*r=NULL)
    {
        data=da;
        lchild=l;
        rchild=r;
        isFirst=true;
    }
    bool isLeaf()const;
    void visit();
};
template<class T>
bool TreeNode<T>::isLeaf()const
{
    if(lchild==NULL && rchild==NULL)
        return true;
    else return false;
}
template<class T>
void TreeNode<T>::visit()
{
    cout<<"访问"<<this->data<<endl;
}
template<class T>
class BinaryTree//class后面不带括号
{
private:
    TreeNode<T>*root;
public:
    BinaryTree();
    ~BinaryTree();
    bool IsEmpty();
    void Distroy(TreeNode<T>*root);
    TreeNode<T>*getRoot(){return root;}
    TreeNode<T>*create();
    void PreOrder(TreeNode<T>*root);
    void InOrder(TreeNode<T>*root);
    void PostOrder(TreeNode<T>*root);
    void FeiPreOrder();
    void FeiInOrder();
    void FeiPostOrder();
};
template<class T>
BinaryTree<T>::BinaryTree()
{
    root=new TreeNode<T>(0);
    root=this->create();
}
template<class T>
BinaryTree<T>::~BinaryTree()
{
    //暂时忽略
}
/*template<class T>
void BinaryTree<T>::Destroy(TreeNode<T>*p)
{
    ;
}*/
template<class T>
bool BinaryTree<T>::IsEmpty()
{
    if(root==NULL)
        return true;
    else
        return false;
}
template<class T>
TreeNode<T>*BinaryTree<T>::create()
{
    root->data='A';
    root->lchild=new TreeNode<T>('B');
    root->rchild=new TreeNode<T>('C');
    root->lchild->lchild=new TreeNode<T>('D');
    root->lchild->rchild=new TreeNode<T>('E');
    root->rchild->lchild=new TreeNode<T>('F');
    root->rchild->rchild=new TreeNode<T>('G');
    return root;
}
template<class T>
void BinaryTree<T>::PreOrder(TreeNode<T>*root)
{
    //cout<<"先序遍历"<<endl;
    if(root!=NULL)
    {
        root->visit();
        PreOrder(root->lchild);
        PreOrder(root->rchild);
    }
}
template<class T>
void BinaryTree<T>::InOrder(TreeNode<T>*root)
{
    //cout<<"中序遍历"<<endl;
    if(root!=NULL)
    {
        InOrder(root->lchild);
        root->visit();
        InOrder(root->rchild);
    }
}
template<class T>
void BinaryTree<T>::PostOrder(TreeNode<T>*root)
{
    //cout<<"后序遍历"<<endl;
    if(root!=NULL)
    {
        PostOrder(root->lchild);
        PostOrder(root->rchild);
        root->visit();
    }
}
template<class T>
void BinaryTree<T>::FeiPreOrder()
{
    stack<TreeNode<T>*>st;
    TreeNode<T>*current=this->root;
    while(!st.empty()||current!=NULL)
    {
        while(current!=NULL)
        {
            current->visit();
            st.push(current);
            current=current->lchild;
        }
        if(!st.empty())
           {
               current=st.top();
               st.pop();
               current=current->rchild;
           }
    }
}
template<class T>
void BinaryTree<T>::FeiInOrder()
{
    stack<TreeNode<T>*>st;
    TreeNode<T>*current=root;
    while(!st.empty()||current!=NULL)
    {
        while(current!=NULL)
        {
            st.push(current);
            current=current->lchild;
        }
        if(!st.empty())
        {
            current=st.top();
            st.pop();
            current->visit();
            current=current->rchild;
        }
    }
}
template<class T>
void BinaryTree<T>::FeiPostOrder()
{
    stack<TreeNode<T>*>st;
    TreeNode<T>*current=root;
    while(!st.empty()||current!=NULL)
    {
        while(current!=NULL)
        {
            st.push(current);
            current=current->lchild;
        }
        if(!st.empty())
        {
            current=st.top();
            if(current->isFirst==true)
            {
                current->isFirst=false;
                current=current->rchild;
            }
            else
            {
                current->visit();
                st.pop();
                current=NULL;//这个里面因为我前面的current被改了，所以不是null了，要恢复null实现从栈里面接着寻找元素
                //current=st.top();
            }
        }
    }
}
int main()
{
    BinaryTree<char> btree;
    cout<<"前序遍历结果"<<endl;
    btree.PreOrder(btree.getRoot());
    cout<<endl;
    btree.FeiPreOrder();
    cout<<"中序遍历结果"<<endl;
    btree.InOrder(btree.getRoot());
    cout<<endl;
    btree.FeiInOrder();
    cout<<"后序遍历结果"<<endl;
    btree.PostOrder(btree.getRoot());
    cout<<endl;
    btree.FeiPostOrder();
    return 0;
}

然后还有一个不修改定义的后序遍历，后序比前中麻烦的地方是需要考虑到各种缺孩子的情况
leetcode上后序遍历已经ac的代码

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        TreeNode*prev;
        TreeNode*current=root;
        stack<TreeNode*>st;
        vector<int>list;
        while(!st.empty()||current!=NULL)
        {
            while(current!=NULL)
            {
                st.push(current);
                prev=current;
                current=current->left;
            }
            if(!st.empty())
            {
                current=st.top();
                if(current->left==NULL && current->right!=NULL)
                {
                    if(prev==current->right)
                    {
                        list.push_back(current->val);
                        st.pop();
                        prev=current;
                        current=NULL;
                    }
                    else
                        current=current->right;
                }
                else if(prev!=current->left||current->right==NULL)
                {
                    list.push_back(current->val);
                    st.pop();
                    prev=current;
                    current=NULL;
                }
                else
                {
                    current=current->right;
                }
            }
        }
        return list;
    }
};

看了一个大佬的文章写的很清晰，复习神器
https://www.cnblogs.com/hapjin/p/5679482.html