Construct Binary Tree from Preorder and Inorder Traversal

题目大意

　　https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

　　从前序和中序序列构建一棵树

题目分析

　　递归即可，这里用了Arrays.copyOfRange方法，最开始没用明白，如果传入参数有误会抛出异常。代码实现还不够优雅，以后有时间再改改

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null) {
            return null;
        }
        if(preorder.length == 0 || inorder.length == 0 || preorder.length != inorder.length) {
            return null;
        }
        int first = preorder[0];
        int loc = -1;
        for(int i = 0; i < inorder.length; i++) {
        	if(inorder[i] == first) {
        		loc = i;
        		break;
        	}
        }
        int[] newIn1 = new int[0];
        if(0 < loc) {
            newIn1 = Arrays.copyOfRange(inorder, 0, loc);
        }
        
        int[] newPre1 = new int[0];
        if(loc > 0 && loc + 1 <= preorder.length) {
            newPre1 = Arrays.copyOfRange(preorder, 1, loc + 1);
        }
        TreeNode left = buildTree(newPre1, newIn1);
        int[] newIn2 = new int[0];
        int[] newPre2 = new int[0];
        if(loc + 1 < inorder.length) {
            newIn2 = Arrays.copyOfRange(inorder, loc + 1, inorder.length);
            newPre2 = Arrays.copyOfRange(preorder, loc + 1, preorder.length);
        }
        TreeNode right = buildTree(newPre2, newIn2);
        TreeNode ret = new TreeNode(first);
        ret.left = left;
        ret.right = right;
        return ret;
    }
}