UVa Problem 112 - Tree Summing

// UVa Problem 112 - Tree Summing
// Verdict: Accepted
// Submission Date: 2011-11-26
// UVa Run Time: 0.304s
//
// 版权所有（C）2011，邱秋。metaphysis # yeah dot net
//
// [解题方法]
// 本题可以归结为数据结构问题。
//
// 使用链表来表示树，若是使用数组，可能因为树的深度较大导致数组过大。关键是如何将输入解析成链表表
// 示的树，本题解利用了 cin.putback() 帮助完成。解析完成，剩下的就是树遍历的问题了。

#include <iostream>
#include <cstdlib>

using namespace std;

struct node
{
struct node *parent;
struct node *leftChild, *rightChild;
int value;
};

bool haveTargetSum, emptyTree;

// 将路径和保存在叶子节点上。
void summingTree(node *current)
{
if (current->leftChild != NULL)
{
current->leftChild->value += current->value;
summingTree(current->leftChild);
}

if (current->rightChild != NULL)
{
current->rightChild->value += current->value;
summingTree(current->rightChild);
}
}

// 利用 cin.putback() 递归解析树。
void parseTree(node *current)
{
bool isLeaf = false;

char c;
while (cin >> c, c != '(')
;

// 需要考虑负数的情况。
cin >> c;
if (isdigit(c) || c == '-')
{
int sign = (c == '-' ? (-1) : 1);
int number;

if (isdigit(c))
number = c - '0';
else
number = 0;

while (cin >> c, isdigit(c))
{
number *= 10;
number += (c - '0');
}

cin.putback(c);
current->value = number * sign;
}
else
{
cin.putback(c);

// 若当前节点为空，则将父节点的相应子节点设为空。
if (current->parent != NULL)
{
if (current == current->parent->leftChild)
current->parent->leftChild = NULL;
else
current->parent->rightChild = NULL;
}
else
emptyTree = true;

isLeaf = true;
}

// 若不是叶子节点，则继续解析子树。
if (!isLeaf)
{
node *left = new node;
current->leftChild = left;
left->parent = current;
parseTree(left);

node *right = new node;
current->rightChild = right;
right->parent = current;
parseTree(right);
}

while (cin >> c, c != ')')
;
}

// 遍历树，检查叶子节点保存的路径和是否为目标值。
void travelTree(node *current, int targetSum)
{
if (haveTargetSum)
return;

if (current->leftChild == NULL && current->rightChild == NULL)
if (current->value == targetSum)
haveTargetSum = true;

if (current->leftChild != NULL)
travelTree(current->leftChild, targetSum);

if (current->rightChild != NULL)
travelTree(current->rightChild, targetSum);
}

int main(int argc, char const *argv[])
{
int targetSum;
while (cin >> targetSum)
{
node *root = new node;

emptyTree = false;
parseTree(root);

summingTree(root);

haveTargetSum = false;
if (!emptyTree)
travelTree(root, targetSum);

cout << (haveTargetSum ? "yes\n" : "no\n");

delete root;
}

return 0;
}