题目

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并输出它的后序遍历序列。

代码

/*---------------------------------------
*   日期：2015-07-20
*   作者：SJF0115
*   题目: 8.重建二叉树
*   结果：AC
*   网址：http://www.nowcoder.com/books/coding-interviews/8a19cbe657394eeaac2f6ea9b0f6fcf6?rp=1
*   来源：剑指Offer
*   博客：
-----------------------------------------*/
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;struct TreeNode{int val;TreeNode *left;TreeNode *right;TreeNode(int x):val(x),left(nullptr),right(nullptr){}
};class Solution{
public:struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {int size = pre.size();if(size == 0){return nullptr;}//ifreturn PreInBuildTree(pre,in,0,0,size);}
private:TreeNode* PreInBuildTree(vector<int> preorder,vector<int> inorder,int preIndex,int inIndex,int size){if(size == 0){return nullptr;}//if// 根节点TreeNode* root = new TreeNode(preorder[preIndex]);// 寻找根节点在中序遍历数组的下标int index = 0;for(int i = 0;i < size;++i){// 注意：inorder[inIndex+i]if(preorder[preIndex] == inorder[inIndex+i]){index = inIndex+i;break;}//if}//for// 左子树个数int leftSize = index - inIndex;// 右子树个数int rightSize = size - leftSize - 1;// 左子树root->left = PreInBuildTree(preorder,inorder,preIndex+1,inIndex,leftSize);// 右子树root->right = PreInBuildTree(preorder,inorder,preIndex+1+leftSize,index+1,rightSize);return root;}
};void PostOrder(TreeNode* root){if(root){PostOrder(root->left);PostOrder(root->right);cout<<root->val<<endl;}//if
}int main(){Solution s;vector<int> preOrder = {1,2,4,7,3,5,6,8};vector<int> inOrder = {4,7,2,1,5,3,8,6};TreeNode* root = s.reConstructBinaryTree(preOrder,inOrder);PostOrder(root);return 0;
}