重建二叉树

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
 
#前序遍历：根 - 左节点 - 右节点
#中序遍历：左节点- 根 - 右节点
#后序遍历：左节点 - 右节点 - 根节点
 
 
class Solution:
    # 返回构造的TreeNode根节点
    def reConstructBinaryTree(self, pre, tin):
        # write code here
        if not pre or not tin:
            return None
        root = TreeNode(pre.pop(0))
        index = tin.index(root.val)
        root.left = self.reConstructBinaryTree(pre,tin[:index])
        root.right = self.reConstructBinaryTree(pre,tin[index+1:])
        return root

 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> vin) {
13         if(pre.empty()||vin.empty()) return NULL;
14         //先建立一个头结点
15         TreeNode* head = new TreeNode(pre[0]);
16         //找到vin的根节点的索引
17         //定义一个索引变量
18         int root_index=0;
19         for(int i=0;i<vin.size();i++){
20             if(vin[i]==pre[0]){
21                 root_index = i;
22                 break;
23             }
24         }
25         vector<int> pre_left,pre_right,vin_left,vin_right;
26         for(int i=0;i<root_index;i++){
27             vin_left.push_back(vin[i]);
28             pre_left.push_back(pre[i+1]);
29         }
30         for(int j=root_index+1;j<pre.size();j++){
31             pre_right.push_back(pre[j]);
32             vin_right.push_back(vin[j]);
33         }
34         head->left = reConstructBinaryTree(pre_left,vin_left);
35         head->right = reConstructBinaryTree(pre_right,vin_right);
36         return head;
37     }
38 };