【day-14】二叉树-剑指offer难题合集

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#include <iostream>#include <vector>#include <map>#include <cstring>#include <sstream>#include <stack>using namespace std;struct TreeNode{    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :        val(x), left(NULL), right(NULL)    {    }};//反序列号二叉树void Serialize(TreeNode *root,string &s){    if(root!=NULL)    {        stringstream sstream;        sstream<<root->val;        string temp;        sstream>>temp;        s=s+temp+',';        Serialize(root->left,s);        Serialize(root->right,s);    }    else    {        s=s+"#,";    }}char * Serialize(TreeNode *root){    string res;    Serialize(root,res);    char * fuck=new char[res.size()+1];    int i=0;    for(; i<res.size(); i++)    {        fuck[i]=res[i];    }    i--;    while(i>=0 && (res[i]==',' || res[i]<='#') )    {        i--;    }    i++;    fuck[i++]='\0';    return fuck;}//序列号二叉树TreeNode* Deserialize(char *str,int &str_index){    if(str==NULL||str_index>=strlen(str))    {        return NULL;    }    if(str[str_index]==',')str_index++;    if(str[str_index]=='#')    {        str_index++;        return NULL;    }    int num=0;    while(str[str_index]>='0'&&str[str_index]<='9')    {        num=num*10+(str[str_index]-'0');        str_index++;    }    TreeNode *root=new TreeNode(num);    root->left=Deserialize(str,str_index);    root->right=Deserialize(str,str_index);    return root;}TreeNode* Deserialize(char *str){    int str_index=0;    TreeNode* root=Deserialize(str,str_index);    return root;}//求二叉树的深度int treeDepth(TreeNode *root){    if(root==NULL)        return 0;    int left=treeDepth(root->left);    int right=treeDepth(root->right);    return (left>right)?left+1:right+1;}//判断二叉树是不是平衡二叉树，平衡二叉树是指在二叉树中任意节点的左右子树的深度不超过1bool isBlance(TreeNode* root){    if(root==NULL)        return true;    int left=treeDepth(root->left);    int right=treeDepth(root->right);    int dif=left-right;    if(dif>1||dif<-1)        return false;    return isBlance(root->left)&&isBlance(root->right);}bool dfs(TreeNode *left,TreeNode * right){    if(left!=NULL && right!=NULL && left->val==right->val)    {        return dfs(left->left,right->right)&&dfs(left->right,right->left);    }    else if(left==NULL && right==NULL)    {        return true;    }    return false;}bool isSymmetrical(TreeNode* root){    if(root==NULL) return true;    return dfs(root->left,root->right);}//二叉树的子结构bool ok(TreeNode* pRoot1, TreeNode* pRoot2){    if(pRoot1==NULL && pRoot2==NULL)    {        return true;    }    else if(pRoot1!=NULL && pRoot2==NULL)    {        return true;    }    else if(pRoot1==NULL && pRoot2!=NULL)    {        return false;    }    if(pRoot1->val==pRoot2->val)    {        return ok(pRoot1->left,pRoot2->left) && ok(pRoot1->right,pRoot2->right);    }    return false;}bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2){    if(pRoot2==NULL) return false;    stack<TreeNode *>s;    TreeNode *pCur=pRoot1;    while(pCur!=NULL ||!s.empty())    {        while(pCur!=NULL)        {            if( ok(pCur, pRoot2) )            {                return true;            }            s.push(pCur);            pCur=pCur->left;        }        pCur=s.top();        s.pop();        pCur=pCur->right;    }    return false;}//二叉树的镜像void Mirror(TreeNode *pRoot){    if(pRoot==NULL) return;    TreeNode *temp;    swap(pRoot->left,pRoot->right);    Mirror(pRoot->left);    Mirror(pRoot->right);}/*    _30_   /    \  10    20 /     /  \50    45  35前序遍历得到下面30,10,50,#,#,#,20,45,#,#,35,#,#    _30_   /    \  10    20 /     /  \50    45  35            \            70              \              55前序遍历得到下面30,10,50,#,#,#,20,45,#,#,35,#,70,#,55*//*    _30_   /    \  10    10 /  \  /  \50  4545  50前序遍历得到下面30,10,50,#,#,45,#,#,10,45,#,#,50,#,#*//*前序遍历preorder traversal  根左右后序遍历postorder traversal 左右跟中序遍历inorder traversal*//*    _20_   /    \  19    28 /     /  \18    25  29            \            50              \              55前序遍历得到下面20,19,18,#,#,#,28,25,#,#,29,#,50,#,55*///前序遍历vector<int> preorder(TreeNode* root){    vector<int> res;    if(root==NULL)    {        return res;    }    stack<TreeNode *> s;    TreeNode *pCur=root;    while(pCur!=NULL||!s.empty())    {        while(pCur!=NULL)        {            res.push_back(pCur->val);            s.push(pCur);            pCur=pCur->left;        }        pCur=s.top();        s.pop();        pCur=pCur->right;    }    return res;}vector<int> inorder(TreeNode * root){    vector<int> res;    if(root==NULL) return res;    stack<TreeNode *> s;    TreeNode * pCur=root;    while(pCur!=NULL || !s.empty())    {        while(pCur!=NULL)        {            s.push(pCur);            pCur=pCur->left;        }        pCur=s.top();        s.pop();        res.push_back(pCur->val);        pCur=pCur->right;    }    return res;}vector<int> postorder(TreeNode* root){    vector<int> res;    if(root==NULL) return res;    /*    一个节点有三个状态State：    1、没有访问过 T.end()    2、去过右边 ‘L’    3、去过右边 ‘R’    */    map<TreeNode*,char> visitState;    stack<TreeNode *>s;    TreeNode* pCur=root;    do    {        while(pCur!=NULL && visitState.find(pCur)==visitState.end())        {            s.push(pCur);            visitState[pCur]='L';            pCur=pCur->left;        }        pCur=s.top();        if(visitState[pCur]=='L')        {            visitState[pCur]='R';            pCur=pCur->right;        }        else if(visitState[pCur]=='R')        {            res.push_back(pCur->val);            s.pop();        }    }    while(!s.empty());    return res;}vector<int> postorder2(TreeNode* root){    vector<int> res;    if(root==NULL) return res;    TreeNode * pCur=NULL, *pLast=NULL;    stack<TreeNode *> s;    pCur=root;    while(pCur!=NULL  )    {        s.push(pCur);        pCur=pCur->left;    }    while(!s.empty())    {        pCur=s.top();        //左右子树为空，或者上次的输出的节点为右        if(pCur->right==NULL ||pCur->right==pLast)        {            res.push_back(pCur->val);            pLast=pCur;            s.pop();        }        else        {            pCur=pCur->right;            while(pCur!=NULL  )            {                s.push(pCur);                pCur=pCur->left;            }        }    }}template<class T>void print_vector(vector<int> input){    for(auto it=input.begin(); it!=input.end(); ++it)    {        cout<<*it<<" ";    }    cout<<endl;}//前序遍历+中序遍历 ==》构建一棵树TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin){    if(pre.size()==0) return NULL;    if(pre.size()==1) return new TreeNode(vin[0]);    int rootVal=pre[0];    TreeNode* root=new TreeNode(rootVal);    vector<int> pre_left,pre_right,vin_right,vin_left;    //切割点的位置    int index=0;    for(int i=0; i<vin.size(); i++)    {        if(rootVal==vin[i])        {            index=i;            break;        }        vin_left.push_back(vin[i]);    }    for(int i=1; i<pre.size()&&i<=index; i++)    {        pre_left.push_back(pre[i]);    }    for(int i=index+1; i<vin.size(); i++)    {        vin_right.push_back(vin[i]);        pre_right.push_back(pre[i]);    }    root->left=reConstructBinaryTree(pre_left,vin_left);    root->right=reConstructBinaryTree(pre_right,vin_right);    return root;}//判断是否是 二叉搜索树的后序遍历bool judge(vector<int>& a, int left, int right){    if(left >= right) return true;    int i = right;    while(i > left && a[i - 1] > a[right]) --i;    for(int j = i - 1; j >= left; --j)    {        if(a[j] > a[right])            return false;    }    return judge(a, left, i - 1) && (judge(a, i, right - 1));}bool VerifySquenceOfBST(vector<int> a){    if(!a.size())        return false;    return judge(a, 0, a.size() - 1);}//把二叉树变成双向链表TreeNode* Convert(TreeNode* pRootOfTree){    TreeNode* Last=NULL, *head=NULL;    stack<TreeNode*> s;    TreeNode *pCur=pRootOfTree;    while( pCur!=NULL || !s.empty())    {        while(pCur!=NULL)        {            s.push(pCur);            pCur=pCur->left;        }        pCur=s.top();        if(Last==NULL)        {            head=pCur;            pCur->left=NULL;        }        else        {            Last->right=pCur;            pCur->left=Last;        }        Last=pCur;        pCur=pCur->right;        s.pop();    }    return head;}int main(){    //char *res="30,10,50,#,#,#,20,45,#,#,35,#,#";    //char *res="30,10,50,#,#,#,20,45,#,#,35,#,70,#,55";    //char *res="30,10,50,#,#,45,#,#,10,45,#,#,50,#,#";    char *res="20,19,18,#,#,#,28,25,#,#,29,#,50,#,55";    //根据序列得到二叉树，返回节点的根    TreeNode* root=Deserialize(res);    //重新序列化，并输出~    res=Serialize(root);    cout<<"重新的生成的序列为:"<<res<<endl;    //求出二叉树的深度    int depth=treeDepth(root);    cout<<"二叉树的深度为:"<<depth<<endl;    bool blance=isBlance(root);    if(blance) cout<<"平衡二叉树"<<endl;    else cout<<"不平衡二叉树"<<endl;    bool symmetrical=isSymmetrical(root);    if(symmetrical) cout<<"对称"<<endl;    else cout<<"不对称"<<endl;    vector<int> preorderTraversal;    preorderTraversal=preorder(root);    cout<<"前序遍历：";    print_vector<int>(preorderTraversal);    vector<int> inorderTraversal;    inorderTraversal=inorder(root);    cout<<"中序遍历：";    print_vector<int>(inorderTraversal);    vector<int> postorderTraversal;    postorderTraversal=postorder(root);    cout<<"后序遍历：";    print_vector<int>(postorderTraversal);    postorderTraversal=postorder2(root);    cout<<"后序遍历：";    print_vector<int>(postorderTraversal);    bool BST=VerifySquenceOfBST(postorderTraversal);    if(BST) cout<<"二叉收索树的后续遍历"<<endl;    else cout<<"不是二叉收索树的后续遍历"<<endl;    TreeNode *reRoot=reConstructBinaryTree(preorderTraversal,inorderTraversal);    res=Serialize(root);    cout<<"重建后-》重新的生成的序列为:"<<res<<endl;    bool sub=HasSubtree(root,root);    if(sub) cout<<"子结构"<<endl;    else cout<<"不是子结构"<<endl;    Mirror(root);    res=Serialize(root);    cout<<"对称后-》重新的生成的序列为:"<<res<<endl;    //-----------------------------    cout<<"----把二叉树转成双向链表-------"<<endl;    res="20,19,18,#,#,#,28,25,#,#,29,#,50,#,55";    TreeNode* root2=Deserialize(res);    TreeNode *link=Convert(root2);    while(link!=NULL){        cout<<link->val<<"->";        link=link->right;    }    cout<<endl;    return 0;}
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