重建二叉树

python： 

# -*- 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

c++

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        if(pre.empty()||vin.empty()) return NULL;
        //先建立一个头结点
        TreeNode* head = new TreeNode(pre[0]);
        //找到vin的根节点的索引
        //定义一个索引变量
        int root_index=0;
        for(int i=0;i<vin.size();i++){
            if(vin[i]==pre[0]){
                root_index = i;
                break;
            }
        }
        vector<int> pre_left,pre_right,vin_left,vin_right;
        for(int i=0;i<root_index;i++){
            vin_left.push_back(vin[i]);
            pre_left.push_back(pre[i+1]);
        }
        for(int j=root_index+1;j<pre.size();j++){
            pre_right.push_back(pre[j]);
            vin_right.push_back(vin[j]);
        }
        head->left = reConstructBinaryTree(pre_left,vin_left);
        head->right = reConstructBinaryTree(pre_right,vin_right);
        return head;
    }
};