Construct Binary Tree from Preorder and Inorder Traversal

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

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

 

思路：递归思想

注意事项：画图，把长度计算正确

JAVA代码：

1. private TreeNode buildTree(int[] preorder,int pstart,int[] inorder,int istart,int length) {
2. if(preorder==null || length==0) return null;  // 判定条件
3. if(length==1) {
4. return new TreeNode(preorder[pstart]);
5. }
6. int tmp = preorder[pstart];
7. TreeNode root = new TreeNode(tmp);
8. int i = 0;
9. for(i=0;i<length;i++) {
10. if(inorder[istart+i] == tmp) {
11. break;
12. }
13. }
14. root.left = buildTree(preorder,pstart+1,inorder,istart,i);  // 注意左侧的长度 i
15. root.right = buildTree(preorder,pstart+i+1,inorder,istart+i+1,length-(i+1)); // 注意右侧的长度，为length-（i+1）
16. return root;
17. }
18. public TreeNode buildTree(int[] preorder, int[] inorder) {
19. if(preorder==null || inorder==null) return null;
20. int plen = preorder.length;
21. int ilen = inorder.length;
22. if(plen != ilen) return null;
23. return buildTree(preorder,0,inorder,0,plen);
24. }