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.

思路：树形结构的中序和后序，找好相对位置，并控制好边界

java代码：

1. TreeNode helper(int[] inorder,int ist,int[] postorder,int pst,int len) {
2. if(len<1) return null;  //无左节点或者右节点
3. if(len==1) {
4. return new TreeNode(inorder[ist]);
5. }
6. int k = pst + len - 1;
7. int tmp = postorder[k];
8. int i=0;
9. for(i=0;i<len;i++) {
10. if(inorder[ist+i] == tmp) break;
11. }
12. TreeNode root = new TreeNode(inorder[ist+i]);
13. root.left = helper(inorder,ist,postorder,pst,i);
14. root.right = helper(inorder,ist+i+1,postorder,pst+i,len-1-i);  //len - 1 - i,已经处理一个节点了
15. return root;
16. }
17. public TreeNode buildTree(int[] inorder, int[] postorder) {
18. int ilen = inorder.length;
19. int plen = postorder.length;
20. if(ilen == 0) return null;
21. if(ilen != plen) return null;
22. return helper(inorder,0,postorder,0,ilen);
23. }