Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

 

和pre & in 是一样的。

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode buildTree(int[] inorder, int[] postorder) {
12         // IMPORTANT: Please reset any member data you declared, as
13         // the same Solution instance will be reused for each test case.
14         return inorder_postorder(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
15     }
16     public TreeNode inorder_postorder(int[] in, int is, int ie, int[] po, int ps, int pe){
17         if(ps > pe || is > ie) return null;
18         TreeNode root = new TreeNode(po[pe]);
19         int ind = 0;
20         for(int i = is; i <= ie; i++)
21             if(in[i] == root.val){
22                 ind = i;
23                 break;
24         }
25         int len = ind - is;
26         root.left = inorder_postorder(in, is, ind - 1,po, ps, ps + len - 1);
27         root.right = inorder_postorder(in, ind + 1, ie, po, ps + len, pe - 1);
28         return root;
29     }
30 }

 

posted on 2013-11-16 06:43  Step-BY-Step  阅读(139)  评论(0)  编辑  收藏  举报

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