剑指offer刷题之c++实现的根据二叉树的前序和中序遍历重建二叉树

#include "myHead.h"#include "allConstructBinaryTree.cpp" /*利用前序和后序遍历，构造二叉树。前序第一个节点就是根节点，然后根据这个节点去后序中找到这个数，并记下这个数的下标作为划分的标准。这个数划分了两个范围。左边一拨全是左子树，右边一拨全是右子树。 可以统计左子树的个数，下一次，在前序中去掉第一个再开始，以这个中序序列的第二个数为左子树的根，继续在中序的左子树一拨中划分左右子树。知道叶子节点。如果划分后序的左子树个数为零，那么只有右子树，如果划分后序的右子树个数为零，那么只有左子树。 */    void create2(TreeNode *&T,vector<int> pre,vector<int> in,int l1,int h1,int l2,int h2);struct TreeNode* reConstructBinaryTreeByPreIn(vector<int> pre,vector<int> in) {int l1,h1,l2,h2;TreeNode *T;        l1 = 0;        h1 = pre.size()-1;        l2 = 0;        h2 = in.size()-1;               printf("first: l1=%d,h1=%d,l2=%d,h2=%d",l1,h1,l2,h2);create2(T,pre,in,l1,h1,l2,h2); inOrder(T);cout<<"return "<<endl;        return T;    }         void create2(TreeNode *&T,vector<int> pre,vector<int> in,int l1,int h1,int l2,int h2){int j=0,l=0;printf("\tl1=%d,h1=%d,l2=%d,h2=%d\n",l1,h1,l2,h2);T = (TreeNode *) malloc(sizeof(TreeNode)) ;T->val = pre[l1];cout<<"pre2:"<<T->val<<endl;        for(j = l2;j<=h2;j++){                        if(pre[l1] == in[j]){            printf("j=%d",j);                break;            }        }         l = j-l2;if(l2 == j){cout<<"left null"; T->left = NULL;}else        create2(T->left,pre,in,l1+1,l1+l,l2,j-1);        if(h2 == j){        cout<<"right null";        T->right = NULL;        }         else        create2(T->right,pre,in,l1+l+1,h1,j+1,h2); }            int main(){    vector<int> pre,pre1;vector<int> in;vector<int> post,post1;pre.push_back(1);pre.push_back(2); pre.push_back(4);pre.push_back(7);pre.push_back(3);pre.push_back(5);pre.push_back(6);pre.push_back(8);in.push_back(4);in.push_back(7);in.push_back(2);in.push_back(1);in.push_back(5);in.push_back(3);in.push_back(8);in.push_back(6); cout<<"pre: ";for(vector<int>::iterator it=pre.begin();it<pre.end();++it){  cout<<*it<<" ";}cout<<endl;cout<<"in:  ";for(vector<int>::iterator it=in.begin();it<in.end();++it){  cout<<*it<<" ";}cout<<endl;TreeNode *t =reConstructBinaryTreeByPreIn(pre,in);cout<<endl<<"finally";    inOrder(t);return 1;}

拓展：根据严格二叉树的前序和后序遍历重建二叉树

#include "myHead.h"#include "allConstructBinaryTree.cpp" /*利用前序和中序，其中给定的序列的数都不重复，从而构造一颗严格二叉树，所谓严格二叉树每颗父节点要不都没有左右孩子 ，要不左右子树都不空但是，显然这个限制条件就不能在普通二叉树中使用了 */void create(TreeNode *&T,vector<int> pre,vector<int> in,int l1,int h1,int l2,int h2);struct TreeNode* reConstructBinaryTreeByPrePostInConditions(vector<int> pre,vector<int> in) {int l1,h1,l2,h2;TreeNode *T;        l1 = 0;        h1 = pre.size()-1;        l2 = 0;        h2 = in.size()-1;               printf("first: l1=%d,h1=%d,l2=%d,h2=%d",l1,h1,l2,h2);        create(T,pre,in,l1,h1,l2,h2);         return T;    }    void create(TreeNode *&T,vector<int> pre,vector<int> in,int l1,int h1,int l2,int h2){    int i = 0,j=0,l=0;printf("l1=%d,h1=%d,l2=%d,h2=%d\n",l1,h1,l2,h2);if(l1<= h1)    {T = (TreeNode *) malloc(sizeof(TreeNode)) ;T->val = pre[l1];            if(l1 == h1){                T->left = NULL;                T->right = NULL;               }              else{                 for(j = l2;j<=h2;j++){                    i++;                    if(pre[l1+1] == in[j]){                        break;                    }                }                i--;l = j-l2+1;                create(T->left,pre,in,l1+1,l1+l,l2,j);                create(T->right,pre,in,l1+l+1,h1,j+1,h2-1);     }    }     }             int main(){    vector<int> pre1;vector<int> post1; pre1.push_back(1); pre1.push_back(2); pre1.push_back(4); pre1.push_back(5); pre1.push_back(3); pre1.push_back(6); pre1.push_back(8); pre1.push_back(9); pre1.push_back(7); pre1.push_back(10); pre1.push_back(11); pre1.push_back(12); pre1.push_back(13); post1.push_back(4); post1.push_back(5); post1.push_back(2); post1.push_back(8);  post1.push_back(9); post1.push_back(6); post1.push_back(10); post1.push_back(12);  post1.push_back(13); post1.push_back(11); post1.push_back(7); post1.push_back(3);  post1.push_back(1);  cout<<"pre1:  ";for(vector<int>::iterator it=pre1.begin();it<pre1.end();++it){  cout<<*it<<" ";}cout<<endl;cout<<"post1:  ";for(vector<int>::iterator it=post1.begin();it<post1.end();++it){  cout<<*it<<" ";}cout<<endl;TreeNode *t =reConstructBinaryTreeByPrePostInConditions(pre1,post1);cout<<endl<<"finally:"<<endl;    inOrder(t);return 1;}