第四十三题 通过前序遍历和中序遍历构建二叉树

题目：通过树的前序遍历和中序遍历来确定树的结构并输出，通过递归的方法

#include <iostream>//#include <boost\shared_ptr.hpp>using namespace std;struct NodeTree{int m_value;NodeTree *m_left;NodeTree *m_right;};NodeTree* maketree(int *pre,int *mid,int length){if (pre==nullptr||mid==nullptr||length<=0){cout<<"input is error"<<endl;return nullptr;}NodeTree *root=new NodeTree();//boost::shared_ptr<NodeTree> root(new NodeTree());root->m_value=*pre;root->m_left=nullptr;root->m_right=nullptr;if (length==1){if (*pre==*mid){return root;}else{cout<<"error"<<endl;return nullptr;}}int *midbeg=mid;while (*midbeg!=root->m_value&&midbeg<mid+length){++midbeg;}if (midbeg==mid+length){cout<<"error root"<<endl;return nullptr;}int rootlength=midbeg-mid;//构建左子树if (rootlength>0){root->m_left=maketree(pre+1,mid,rootlength);}//构建右子树if (rootlength<length-1){root->m_right=maketree(pre+rootlength+1,midbeg+1,length-1-rootlength);}return root;}void PrintTreeNode(NodeTree* pNode){if(pNode != nullptr){printf("value of this node is: %d\n", pNode->m_value);if(pNode->m_left != nullptr)printf("value of its left child is: %d.\n", pNode->m_left->m_value);elseprintf("left child is null.\n");if(pNode->m_right != nullptr)printf("value of its right child is: %d.\n", pNode->m_right->m_value);elseprintf("right child is null.\n");}else{printf("this node is null.\n");}printf("\n");}void PrintTree(NodeTree* pRoot){PrintTreeNode(pRoot);if(pRoot != nullptr){if(pRoot->m_left != nullptr)PrintTree(pRoot->m_left);if(pRoot->m_right != nullptr)PrintTree(pRoot->m_right);}}int main(){const int length = 5;int preorder[length] = {1, 2, 3, 4, 5};int inorder[length] = {5, 4, 3, 2, 1};NodeTree *root=nullptr;root=maketree(preorder,inorder,length);PrintTree(root);delete root;root=nullptr;return 0;}

解决