重建二叉树

题目描述

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}，则重建二叉树并返回。

 

基本思想：在中序序列中找到根节点的位置，根结点左边是左子树，右边是右子树。然后分别对左子树和右子树进行递归处理。

在打印树的函数上，我的实现方法是广度优先遍历，也就是按层打印二叉树。

 

#include <iostream>
#include <algorithm>
#include "string.h"
#include "stdio.h"
#include <vector>
#include <deque>
#include <stack>
#include <queue>
#include<map>
#include<utility>
#include "math.h"
using namespace std;

 struct TreeNode {
      int val;
      TreeNode *left;
      TreeNode *right;
      TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Tree{
public:
    void PrintTree(TreeNode* pRoot)
    {
        if(pRoot == NULL)
            return;
        int level = 0;
        int index = 0;
        int globle = 0;
        queue<TreeNode*> queue;

        queue.push(pRoot);
        while(!queue.empty())
        {
            TreeNode* pNode = queue.front();
            cout<<pNode->val<<" ";
            queue.pop();
            if(pNode->left)
            {
                queue.push(pNode->left);
            }
            if(pNode->right)
            {
               queue.push(pNode->right);
            }
        }
    }
};
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        if(pre.size() == NULL || pre.size() == NULL || pre.size()!=vin.size())
            return NULL;

        vector<int> pre_left;
        vector<int> pre_right;
        vector<int> vin_left;
        vector<int> vin_right;
        int index = 0;

        for(int i=0;i<vin.size();i++)
        {
            if(vin[i] == pre[0])
            {
                index = i;
                break;
            }
        }
        for(int i=0;i<vin.size();i++)
        {
            if(i<index)
            {
                vin_left.push_back(vin[i]);
                pre_left.push_back(pre[i+1]);
            }
            if(i>index)
            {
                vin_right.push_back(vin[i]);
                pre_right.push_back(pre[i]);
            }
        }
        TreeNode* pRoot = new TreeNode(pre[0]);
        pRoot->left = reConstructBinaryTree(pre_left,vin_left);//递归左子树
        pRoot->right = reConstructBinaryTree(pre_right,vin_right);//递归右子树

        return pRoot;

    }
};

int main()
{
    vector<int> pre;
    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);

    vector<int> vin;
    vin.push_back(4);
    vin.push_back(7);
    vin.push_back(2);
    vin.push_back(1);
    vin.push_back(5);
    vin.push_back(3);
    vin.push_back(8);
    vin.push_back(6);

    Solution solution;
    Tree tree;
    TreeNode* p = solution.reConstructBinaryTree(pre,vin);
    tree.PrintTree(p);

}