根据一棵树的中序遍历与后序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
    }
    public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
                               int[] postorder, int postLeft, int postRight) {
        // 没有元素了
        if (inRight - inLeft < 1) {
            return null;
        }
        // 只有一个元素了
        if (inRight - inLeft == 1) {
            return new TreeNode(inorder[inLeft]);
        }
        // 后序数组postorder里最后一个即为根结点
        int rootVal = postorder[postRight - 1];
        TreeNode root = new TreeNode(rootVal);
        int rootIndex = 0;
        // 根据根结点的值找到该值在中序数组inorder里的位置
        for (int i = inLeft; i < inRight; i++) {
            if (inorder[i] == rootVal) {
                rootIndex = i;
                break;
            }
        }
        // 根据rootIndex划分左右子树
        root.left = buildTree1(inorder, inLeft, rootIndex,
                postorder, postLeft, postLeft + (rootIndex - inLeft));
        root.right = buildTree1(inorder, rootIndex + 1, inRight,
                postorder, postLeft + (rootIndex - inLeft), postRight - 1);
        return root;
    }
}