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.

本来想先对inorder array做预处理存上val对应的index，结果发现 val可能有duplicates。

/**duplicates
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        return preorder_inorder(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }
    public TreeNode preorder_inorder(int[] pre, int ps, int pe, int[] in, int is, int ie){
        if(ps > pe || is > ie) return null;
        TreeNode root = new TreeNode(pre[ps]);
        int ind = 0;
        for(int i = is; i <= ie; i++)
            if(in[i] == root.val){
                ind = i;
                break;
        }
        int len = ind - is;
        root.left = preorder_inorder(pre, ps + 1, ps + len, in, is, ind - 1);
        root.right = preorder_inorder(pre, ps + 1 + len, pe, in, ind + 1, ie);
        return root;
    }
}