【leetcode刷题笔记】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.

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类似http://www.cnblogs.com/sunshineatnoon/p/3854935.html

只是子树的前序和中序遍历序列分别更新为：

//左子树：

left_prestart = prestart+1

left_preend = prestart+index-instart

//右子树

right_prestart = prestart+index-instart+1

right_preend =  preend

代码如下：

 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 int InorderIndex(int[] inorder,int key){

12         if(inorder == null || inorder.length == 0)

13             return -1;

14         

15         for(int i = 0;i < inorder.length;i++)

16             if(inorder[i] == key)

17                 return i;

18         

19         return -1;

20     }

21     public TreeNode buildTreeRec(int[] preoder,int[] inorder,int prestart,int preend,int instart,int inend){

22         if(instart > inend)

23             return null;

24         TreeNode root = new TreeNode(preoder[prestart]);

25         int index = InorderIndex(inorder, root.val);

26         root.left = buildTreeRec(preoder, inorder, prestart+1, prestart+index-instart, instart, index-1);

27         root.right = buildTreeRec(preoder, inorder, prestart+index-instart+1, preend, index+1, inend);

28         

29         return root;

30     }

31     public TreeNode buildTree(int[] preorder, int[] inorder) {

32         return buildTreeRec(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);

33     }

34 }