LeetCode 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.

 

后序遍历的最后一个元素就是根元素，由于没有重复，就在中序遍历的数组中查找根元素，这样就分成的两段，然后递归。

 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         int length=inorder.length;
13         if (length==0) return null;
14         int rootnumber=postorder[length-1];
15         TreeNode root = new TreeNode(rootnumber);
16 
17         if (length==1) return root;
18         int indexRoot = 0;
19         for (int i = 0; i < length; i++) {
20             if (inorder[i]==rootnumber){
21                 indexRoot=i;
22                 break;
23             }
24         }
25         int[] left=new int[indexRoot];
26         int[] leftPost=new int[indexRoot];
27         int[] right=new int[length-indexRoot-1];
28         int[] rightPost=new int[length-indexRoot-1];
29         for (int i = 0; i < indexRoot; i++) {
30             left[i]=inorder[i];
31             leftPost[i]=postorder[i];
32         }
33         for (int i = indexRoot+1; i <length ; i++) {
34             right[i-1-indexRoot]=inorder[i];
35             rightPost[i-1-indexRoot]=postorder[i-1];
36         }
37         root.left=buildTree(left,leftPost);
38         root.right=buildTree(right,rightPost);
39         return root;
40     }
41 }

 

posted @ 2014-11-10 23:29  birdhack  阅读(108)  评论(0)  编辑  收藏  举报