leetcode------Construct Binary Tree from Preorder and Inorder Traversal

标题：	Construct Binary Tree from Preorder and Inorder Traversal
通过率：	26.5
难度：	中等

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

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

根据前序遍历和中序遍历构建二叉树：看一棵树如下:

    1

   / \

  2   4

   \   \

   5    6
前序为：12546
中序为：25146

从前序得知第一个数一定是树的root，通过中序得到右树和左树

递归调用即可，其中要注意递归的条件个字符串拷贝时的用法，

具体看代码

 

 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[] preorder, int[] inorder) {

12         if(preorder.length==0||inorder.length==0){

13             return null;

14         }

15         TreeNode root=new TreeNode(preorder[0]);

16         int i=0;

17         for(;i<inorder.length;i++){

18             if(inorder[i]==preorder[0])break;

19         }

20         int [] new_pre_left,new_pre_right,new_in_left,new_in_right;

21         if(i<preorder.length){

22             new_in_left=new int[i];

23             System.arraycopy(inorder, 0, new_in_left, 0, i); 

24             new_pre_left=new int[i];

25             System.arraycopy(preorder, 1, new_pre_left, 0, i);  

26             root.left=buildTree(new_pre_left,new_in_left);

27             

28             new_in_right=new int[preorder.length-i-1];

29             System.arraycopy(inorder, i+1, new_in_right, 0, preorder.length-i-1); 

30             new_pre_right=new int[preorder.length-i-1];

31              System.arraycopy(preorder, i+1, new_pre_right, 0, preorder.length-i-1); 

32              root.right=buildTree(new_pre_right,new_in_right);

33             

34         }

35         return root;

36     }

37 }