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

　　思路：前序遍历的第一个为根节点，在中序遍历中查找根节点，划分左右子树，再递归求解

class Solution {
public:
    int findMid(vector<int> &vin, int begin, int end, int number)
    {
        int idx;
        for(idx=begin; idx<=end; ++idx)
        {
            if(vin[idx]==number)break;
        }
        return idx;
    }
    void reBuildTree(TreeNode **proot, vector<int> &pre, int preStart, int preEnd, vector<int> &vin, int vinStart, int vinEnd)
    {
        if(preStart>preEnd || vinStart>vinEnd)return;
        else{
            (*proot)=new TreeNode(pre[preStart]);
            int mid=findMid(vin, vinStart, vinEnd, pre[preStart]);//在中序数组中找到分界点
            reBuildTree(&((*proot)->left), pre, preStart+1, preStart+(mid-vinStart), vin, vinStart, mid-1);//构建左子树
            reBuildTree(&((*proot)->right), pre, preStart+(mid-vinStart)+1, preEnd, vin, mid+1, vinEnd);//构建右子树
        }
    }
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        TreeNode *root=NULL;
        reBuildTree(&root, pre, 0, pre.size()-1, vin, 0, vin.size()-1);//传入二级指针
        return root;
    }
};

 

class Solution {
public:
int getMid(vector<int> &vin, int begin, int end, int target)
{
    while(vin[begin]!=target && begin<=end)++begin;
    return begin;
}
TreeNode * reBuildTree(vector<int> &pre, int &i, vector<int> &vin, int begin, int end)
{
    if(begin>end)return NULL;
    int mid=getMid(vin, begin, end, pre[i]);
    TreeNode * node=new TreeNode(pre[i++]);
    node->left=reBuildTree(pre, i, vin, begin, mid-1);
    node->right=reBuildTree(pre, i, vin, mid+1, end);
    return node;
}
TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin)
{
    if(pre.size()==0 || vin.size()==0 || pre.size()!=vin.size())return NULL;
    int i=0;
    return reBuildTree(pre, i, vin, 0, vin.size()-1);
}
};