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。

pre={1,2,4,7,3,5,6,8},vin={4,7,2,1,5,3,8,6},首先根节点pHead=new treenode(pre[0]);不断的构建左子树，右子树即可。

首先在vin中找出根节点1，那么左子树4,7,2}右子树{5,3,8,6},因此preleft={2,4,7}vinleft={4,7,2},preright={3,5,8,6},vinright={5,3,8,6}

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        
        int n=pre.size();
        int i,pos;
        if(n==0) return NULL;
        vector<int> preleft,preright,vinleft,vinright;
        TreeNode* pHead=new TreeNode(pre[0]);
        for(i=0;i<n;i++)
        {
            if(vin[i]==pre[0])
            {
                pos=i;
                break;
            }
        }
        for(i=1;i<=pos;i++)                //构建左子树
        {
            preleft.push_back(pre[i]);
            vinleft.push_back(vin[i-1]);
        }
        for(i=pos+1;i<n;i++)                //构建右子树
        {
            preright.push_back(pre[i]);
            vinright.push_back(vin[i]);
        }
        pHead->left=reConstructBinaryTree(preleft,vinleft);
        pHead->right=reConstructBinaryTree(preright,vinright);
        return pHead;


    }
};