剑指offer-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。

AC代码

#include <iostream>
#include <vector>
using namespace std;

typedef struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x)
    {
        val = x;
        left = NULL;
        right = NULL;
    }

} TreeNode;

TreeNode *fun(int *pre, int *mid, int len)
{
    if (len == 0)
        return NULL;
    int index;
    TreeNode *p = new TreeNode(pre[0]);
    for (int i = 0; i < len; i++)
    {
        if (pre[0] == mid[i])
        {
            index = i;
            break;
        }
    }
    p->left = fun(pre + 1, mid, index);
    p->right = fun(pre + index + 1, mid + index + 1, len - index - 1);
    return p;
}
// TreeNode *head;

class Solution
{
public:
    TreeNode *reConstructBinaryTree(vector<int> pre, vector<int> vin)
    {

        if (pre.size() == 0 || vin.size() == 0)
            return nullptr;
        int *pre_tree = new int[pre.size()];
        int *vin_tree = new int[vin.size()];
        for (int i = 0; i < pre.size(); i++)
        {
            pre_tree[i] = pre[i];
            vin_tree[i] = vin[i];
        }
        return fun(pre_tree, vin_tree, pre.size());
                // return fun(pre, vin, 0, pre.size(), 0,vin.size());
    }
};

void print(TreeNode *root)
{ //后序遍历
    if (root != NULL)
    {
        print(root->left);
        print(root->right);
        cout << root->val;
    }
}
int main()
{
    Solution a;
    int Tree_pre[8] = {1,2,4,7,3,5,6,8};
    int Tree_vin[8] = {4,7,2,1,5,3,8,6};


    vector<int>pre(Tree_pre, Tree_pre+8); //用数组初始化向量
    vector<int>vin(Tree_vin, Tree_vin+8);

    TreeNode *p = a.reConstructBinaryTree(pre, vin);
    print(p);
    return 0;
}

总结

• 感觉这个oj好麻烦，写个函数的参数都要严格要求
• 对于二叉树的构建这只是个简单的题目=-=，好像没啥总结的。。

来源：CSDN

作者：chenchenxiaojian

链接：https://blog.csdn.net/weixin_42100456/article/details/103916254