剑指offer系列1：重建二叉树

剑指offer第6题：重建二叉树

#include<iostream>
#include<vector>
#include<stack>
using namespace std;
struct TreeNode {
       int val;
       TreeNode *left;
       TreeNode *right;
       
       TreeNode(int x) :
              val(x), left(NULL),right(NULL) {
       }  
       static void PreOrder(TreeNode* root)
       {
           if (root == NULL)                                                                                    //开始忘记写节点为空返回，报错了。这个一定要记住！
           {
               return;
           }
           cout << root->val;
           PreOrder(root->left);
           PreOrder(root->right);
       }
       static void InOrder(TreeNode* root)
       {
           if (root == NULL)
           {
               return;
           }
           InOrder(root->left);
           cout << root->val;
           InOrder(root->right);
       }
 };
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin)
    {
        if (pre.size() != vin.size())
        {
            cout << "length not equal" << endl;
            return NULL;
        }
        if (pre.size() == 0)
        {
            cout << "length=0" << endl;
            return NULL;
        }
        int rootval = pre[0];
        TreeNode *TreeRoot = new TreeNode(rootval);
        int length = pre.size();
        int sig = 0;
        for (int i = 0; i < length; i++)
        {
            if (vin[i] == rootval)
            {
                sig = i;
                break;
            }
        }
        int lengthOfLeft = sig;
        int lengthOfRight = length-sig-1;
        vector<int> leftpre(lengthOfLeft), leftin(lengthOfLeft);
        vector<int> rightpre(lengthOfRight), rightin(lengthOfRight);
        /*                                                                                                         //将左子树和右子树一起建立
            for (int i = 0; i < length; i++)
            {

                if (i < sig)
                {
                    leftpre[i] = pre[i + 1];
                    leftin[i] = vin[i];
                }
                if (i > sig) 
                {

                    for (int i = sig + 1, j = 0; i < length, j < lengthOfRight; i++, j++)
                    {
                        rightpre[j] = pre[i];
                        rightin[j] = vin[i];
                    }
                }
            }
            */
        if (lengthOfLeft > 0)                                                                                       //分开建立，建立左子树
        {
            for (int i = 0; i < lengthOfLeft; i++)
            {
                leftpre[i] = pre[i + 1];
                leftin[i] = vin[i];
            }
        }
        if (lengthOfRight > 0)                                                                                      //建立右子树
        {
            for (int i = sig+1,j=0; i < length, j < lengthOfRight; i++,j++)
            {
                rightpre[j] = pre[i];
                rightin[j] = vin[i];
            }
        }
    


        TreeRoot->left = reConstructBinaryTree(leftpre, leftin);
        TreeRoot->right = reConstructBinaryTree(rightpre, rightin);

        return TreeRoot;

    }
};
int main()
{
    int pre[] = { 1,2,4,7,3,5,6,8 };
    int in[] = { 4,7,2,1,5,3,8,6 };
    vector<int> preorder(pre, pre + 8);
    vector<int> inorder(in, in + 8);
    Solution so;
    TreeNode * root= so.reConstructBinaryTree(preorder, inorder);
    cout << "PreOrder";
    TreeNode::PreOrder(root);
    cout << endl;

    cout << "InOrder ";
    TreeNode::InOrder(root);
    cout << endl;

    return 0;
}

这道题不难，但是做的时候还是有一些细节问题。

链表是随机存储的增删简单的数组，而树是加了一个指针的链表。

对于vector的使用还要多加练习。