LeetCode Construct Binary Tree from Inorder and Postorder Traversal

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

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

后序遍历的最后一个元素就是根元素，由于没有重复，就在中序遍历的数组中查找根元素，这样就分成的两段，然后递归。

/**
 * 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[] inorder, int[] postorder) {
        int length=inorder.length;
        if (length==0) return null;
        int rootnumber=postorder[length-1];
        TreeNode root = new TreeNode(rootnumber);

        if (length==1) return root;
        int indexRoot = 0;
        for (int i = 0; i < length; i++) {
            if (inorder[i]==rootnumber){
                indexRoot=i;
                break;
            }
        }
        int[] left=new int[indexRoot];
        int[] leftPost=new int[indexRoot];
        int[] right=new int[length-indexRoot-1];
        int[] rightPost=new int[length-indexRoot-1];
        for (int i = 0; i < indexRoot; i++) {
            left[i]=inorder[i];
            leftPost[i]=postorder[i];
        }
        for (int i = indexRoot+1; i <length ; i++) {
            right[i-1-indexRoot]=inorder[i];
            rightPost[i-1-indexRoot]=postorder[i-1];
        }
        root.left=buildTree(left,leftPost);
        root.right=buildTree(right,rightPost);
        return root;
    }
}