Given inorder and postorder traversal of a tree, construct the binary tree.
题目大意：给定一个二叉树的中序和后续序列，构建出这个二叉树。
解题思路：首先后序序列的最后一个是根节点，然后在中序序列中找到这个节点，中序序列中这个节点左边的是根节点的左子树，右边的是右子树，由此递归构建出完整的树。
Talk is cheap:
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null) {
return null;
}
int inLen = inorder.length;
int postLen = postorder.length;
if ((inLen == 0 && postLen == 0) || inLen != postLen) {
return null;
}
TreeNode root = new TreeNode(postorder[postLen - 1]);
if (inLen == 1) {
return root;
}
int pos = 0;
for (int i = 0; i < inLen; i++) {
if (inorder[i] == postorder[postLen - 1]) {
pos = i;
break;
}
}
int[] inLeft = Arrays.copyOfRange(inorder, 0, pos);
int[] inRight = Arrays.copyOfRange(inorder, pos + 1, inLen);
int[] postLeft = Arrays.copyOfRange(postorder, 0, pos);
int[] postRight = Arrays.copyOfRange(postorder, pos, postLen - 1);
root.left = buildTree(inLeft, postLeft);
root.right = buildTree(inRight, postRight);
return root;
}