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;
}