[剑指Offer] 4.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】递归,先找到根结点,再找到左右子树的前序与中序序列进行递归

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {
13         
14         if(pre.size() == 0)
15             return NULL;
16         TreeNode* Root = new TreeNode(pre[0]);
17         
18         vector<int> pre1,pre2;
19         vector<int> in1,in2;
20         int root = pre[0];
21         int root_id = 0;
22         
23         //找到根结点在中序序列中位置root_id
24         for(int i = 0;i < in.size();i ++){
25             if(root == in[i]){
26                 root_id = i;
27             }
28         }
29         
30         for(int i = 0;i < root_id;i ++)
31             in1.push_back(in[i]);
32         for(int i = root_id + 1;i < in.size();i ++)
33             in2.push_back(in[i]);
34         
35         for(int i = 1;i < in1.size() + 1;i ++)
36             pre1.push_back(pre[i]);
37         for(int i = in1.size() + 1;i < pre.size();i ++)
38             pre2.push_back(pre[i]);
39         
40         Root->left = reConstructBinaryTree(pre1,in1);
41         Root->right = reConstructBinaryTree(pre2,in2);
42         
43         return Root;
44         
45     }
46 };

 

posted @ 2017-02-24 01:58  Strawberry丶  阅读(173)  评论(0编辑  收藏  举报