二叉树的对称和旋转的问题,如何利用C语言实现这个问题的解答

Problem Description
Uncle Fang is learning Splay. When he heard about the rotate operation, he would like to play a game with Captain Chen. There is a tree with n nodes initially, and the node 1 is the root. The i-th nodes gets a weight of wi, and an attribute of interesting value defined as the sum of the weight of it’s descendants. Following are two operations in this tree:

1.do a rotation on a node x. The rotation is the common rotation in splay tree, it can be right-rotation or left-rotation, as demonstrated in the following picture.If it's impossible to rotate, just ignore it!

2.ask Captain Chen the product of interesting values of the nodes in a subtree rooted at node x.

Input
There are several cases.The first line of the input contains a single integer T (T <= 30) which is the number of test cases.Then comes the T test cases .

For each case, the first line contains two integer numbers n and m(1<=n, m<=100000).

The following n lines describe the tree. In the i-th line, there was three number wi, xi, yi,(0<=w<=100000, 0<=x, y<=n) indicate the weight of node i, the left child of node i, and the right child of node i. If xi or yi equals to 0, it means node i has no such child.

And the following m lines describe the operations. There are two numbers p x in each line. If p is 0, it means do right-rotation on node x; If p is 1, it means do left-rotation on node x; if p is 2, it means ask the product of interesting values of the nodes in a subtree rooted at node x.

Output
For the k-th test case, first output “Case #k:” in a separate line.Then for each query,output a single num in each line after mod 1000000007.

Sample Input
1
3 4
1 2 3
1 0 0
1 0 0
2 1
0 1
2 2
2 1

Sample Output
Case #1:
3
6
2

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