根據(jù)一棵樹的中序遍歷與后序遍歷構(gòu)造二叉樹。
注意:
你可以假設(shè)樹中沒有重復(fù)的元素。
例如，給出
中序遍歷 inorder = [9,3,15,20,7]
后序遍歷 postorder = [9,15,7,20,3]
返回如下的二叉樹：

    3
   / \
  9  20
    /  \
   15   7

代碼

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder.length != postorder.length) {
            return null;
        } 
        
        return buildMyTree(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    
    public TreeNode buildMyTree(int[] inorder,
                                int inStart,
                                int inEnd,
                                int[] postorder,
                                int postStart,
                                int postEnd) {
        if (inStart > inEnd) {
            return null;
        }
        if (postStart > postEnd) {
            return null;
        }
        
        int value = postorder[postEnd];
        int position = findPosition(inorder, inStart, inEnd, value);
        if (position == -1) {
            return null;
        }
        
        TreeNode root = new TreeNode(value);
        root.left = buildMyTree(inorder, inStart, position-1, postorder, postStart, postStart+position-1-inStart);
        root.right = buildMyTree(inorder, position+1, inEnd, postorder, postStart+position-inStart, postEnd-1);
        
        return root;
    }
    
    public int findPosition(int[] inorder, int inStart, int inEnd, int key) {
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == key) {
                return i;
            }
        }
        return -1;
    }
}