题目如下：

Tree Summing

Background

LISP was one of the earliest high-level programming languages and, withFORTRAN, is one of the oldest languages currently being used. Lists,which are the fundamental data structures in LISP, can easily be adaptedto represent other important data structures such as trees.

This problem deals with determining whether binary trees represented asLISP S-expressions possess a certain property.

The Problem

Given a binary tree of integers, you are to write a program thatdetermines whether there exists a root-to-leaf path whosenodes sum to a specified integer. For example, in the tree shown belowthere are exactly four root-to-leaf paths.The sums of the paths are 27, 22, 26, and 18.

Binary trees are represented in the input file as LISP S-expressionshaving the following form.

 
empty tree 		 ::= 		 ()

tree 		 ::= 		 empty tree    (integer tree tree)

The tree diagrammed above is represented by the expression(5 (4 (11 (7 () ()) (2 () ()) ) ()) (8 (13 () ()) (4 () (1 () ()) ) ) )

Note that with this formulation all leaves of a tree are of the form(integer () () )

Since an empty tree has no root-to-leaf paths, any query as to whether apath exists whose sum is a specified integer in an empty treemust be answered negatively.

The Input

The input consists of a sequence of test cases in theform of integer/tree pairs. Each test caseconsists of an integer followed by one or more spaces followed by abinary tree formatted as an S-expression as described above. Allbinary tree S-expressions will be valid, but expressions may bespread over several lines and may contain spaces.There will be one or more test cases in an input file, and input isterminated by end-of-file.

The Output

There should be one line of output for each test case (integer/treepair) in the input file. For each pair I,T (I represents theinteger, T represents the tree) the output is the string yes ifthere is a root-to-leaf path in T whose sum is I and no ifthere is no path in T whose sum is I.

Sample Input

22 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
20 (5(4(11(7()())(2()()))()) (8(13()())(4()(1()()))))
10 (3 
     (2 (4 () () )
        (8 () () ) )
     (1 (6 () () )
        (4 () () ) ) )
5 ()

Sample Output

yes
no
yes
no

这道题我是先建树，再遍历。难点在于建树，因为输入比较麻烦，可以分几行输入，所以对空格要忽略，只有对空树的时候才忽略前导‘（’，若非空树，每遇到一个‘（’代表一个节点成功输入。

AC的代码如下；

UVA Tree Summing,布布扣,bubuko.com

UVA Tree Summing

原文：http://blog.csdn.net/u013840081/article/details/26957901