剑指Offer 33 二叉搜索树的后序遍历序列

输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。

 1 # -*- coding:utf-8 -*-
 2 class Solution:
 3     def compareTree(self,inOrder,sequence):
 4         n = len(sequence)
 5         if n == 0:
 6             return True
 7         if n == 1:
 8             return True if inOrder[0] == sequence[0] else False
 9         root1 = sequence[-1]
10         for i in range(n):
11             if inOrder[i] == root1:
12                 break
13         in1 = inOrder[:i]
14         last1 = sequence[:i]
15         
16         in2 = inOrder[i+1:]
17         last2 = sequence[i:n-1]
18         return self.compareTree(in1,last1) and self.compareTree(in2,last2)
19             
20         
21     def VerifySquenceOfBST(self, sequence):
22         n = len(sequence)
23         if n == 0:
24             return False
25         inOrder = sorted(sequence)
26         root1 = sequence[-1]
27         for i in range(n):
28             if inOrder[i] == root1:
29                 break
30         in1 = inOrder[:i]
31         last1 = sequence[:i]
32         
33         in2 = inOrder[i+1:]
34         last2 = sequence[i:n-1]
35         return self.compareTree(in1,last1) and self.compareTree(in2,last2)
36         # write code here

Java版代码，leetcode地址：

 1 class Solution {
 2     public boolean compareTree(int[] inorder, int[] postorder) {
 3         int n = postorder.length;
 4         if (n == 0) {
 5             return true;
 6         }
 7         if (n == 1) {
 8             return inorder[0] == postorder[0];
 9         }
10         int root1 = postorder[n - 1];
11         int i = 0;
12         for (int j = 0; j < n; j++) {
13             if (inorder[j] == root1) {
14                 i = j;
15                 break;
16             }
17         }
18 
19         int[] in1 = Arrays.copyOfRange(inorder, 0, i);
20         int[] last1 = Arrays.copyOfRange(postorder, 0, i);
21 
22         int[] in2 = Arrays.copyOfRange(inorder, i + 1, n);
23         int[] last2 = Arrays.copyOfRange(postorder, i, n - 1);
24 
25         return compareTree(in1, last1) && compareTree(in2, last2);
26     }
27 
28     public boolean verifyPostorder(int[] postorder) {
29         int n = postorder.length;
30         if (n == 0) {
31             return true;//leetoce的判断，空数组返回true
32         }
33         int[] inorder = Arrays.copyOf(postorder, n);
34         Arrays.sort(inorder);
35         int root1 = postorder[n - 1];
36         int i = 0;
37         for (int j = 0; j < n; j++) {
38             if (inorder[j] == root1) {
39                 i = j;
40                 break;
41             }
42         }
43         int[] in1 = Arrays.copyOfRange(inorder, 0, i);
44         int[] last1 = Arrays.copyOfRange(postorder, 0, i);
45 
46         int[] in2 = Arrays.copyOfRange(inorder, i + 1, n);
47         int[] last2 = Arrays.copyOfRange(postorder, i, n - 1);
48 
49         return compareTree(in1, last1) && compareTree(in2, last2);
50     }
51 }

思路分析：二叉树搜索树的中序遍历是有序的。因此对于一个二叉树的后序遍历顺序进行排序，就可以得到其中序遍历的结果。

使用中序遍历和后续遍历可以确定一棵二叉树。

posted on 2019-06-13 08:06 Sempron2800+ 阅读(205) 评论(0) 编辑收藏举报