剑指Offer 07 重建二叉树

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。

 1 # -*- coding:utf-8 -*-
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 class Solution:
 8     # 返回构造的TreeNode根节点
 9     def reConstructBinaryTree(self, pre, tin):
10         n = len(pre)
11         if n == 0:
12             return None
13         if n == 1:
14             return TreeNode(pre[0])
15         root = TreeNode(pre[0])
16         for i in range(len(tin)):
17             if tin[i] == pre[0]:
18                 break
19         root.left = self.reConstructBinaryTree(pre[1:i+1],tin[:i])
20         root.right = self.reConstructBinaryTree(pre[i+1:],tin[i+1:])
21         return root
22         # write code here

补充Java的实现，leetcode地址：

 1 class Solution {
 2     public TreeNode buildTree(int[] preorder, int[] inorder) {
 3         int n = preorder.length;
 4         if (n == 0) {
 5             return null;
 6         }
 7         if (n == 1) {
 8             return new TreeNode(preorder[0]);
 9         }
10         TreeNode root = new TreeNode(preorder[0]);
11         int len = inorder.length;
12         int i = 0;
13         while (i < len) {
14             if (inorder[i] == preorder[0]) {
15                 break;
16             }
17             i++;
18         }
19         int[] newpre1 = Arrays.copyOfRange(preorder, 1, i + 1);
20         int[] newin1 = Arrays.copyOfRange(inorder, 0, i);
21         root.left = this.buildTree(newpre1, newin1);
22 
23         int[] newpre2 = Arrays.copyOfRange(preorder, i + 1, preorder.length);
24         int[] newin2 = Arrays.copyOfRange(inorder, i + 1, inorder.length);
25         root.right = this.buildTree(newpre2, newin2);
26 
27         return root;
28     }
29 }

posted on 2019-06-12 22:01 Sempron2800+ 阅读(95) 评论(0) 编辑收藏举报