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 }