129. Sum Root to Leaf Numbers

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本文探讨了一道经典的二叉树遍历问题——求所有从根节点到叶子节点的路径数值之和。通过递归算法实现了路径求和，并给出了完整的C++实现代码。

129. Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

Code：

#include<iostream>
using namespace std;
class node {
public:
	int num;
	//int sum;
	node * left;
	node * right;
	node() {}
	node(int n) { this->num = n; }
};

void insert(node * t, int num)
{
	node *corrent = t;
	if (corrent->num < num&&corrent->right != NULL)
		insert(corrent->right, num);

	else if (corrent->num > num&&corrent->left != NULL)
		insert(corrent->left, num);
	else
	{
		if (corrent->num < num)
			corrent->right = new node(num);
		else if (corrent->num > num)
			corrent->left = new node(num);
		return;
	}

}

/*
void bianli(node *t,int &sum)
{
node *corrent = t;
if (corrent->left == NULL && corrent->right == NULL)
{
cout << corrent->sum << endl;
sum += corrent->sum;
return;
}
if (corrent->left != NULL)
{
corrent->left->sum += corrent->sum*10;
bianli(corrent->left,sum);

}
if (corrent->right != NULL)
{
corrent->right->sum += corrent->sum*10;
bianli(corrent->right,sum);
}
}
*/

int bianli(node *t,int pre)
{
	if (t == NULL)
		return 0;
	else 
	{
		int tmp = pre * 10 + t->num;
		if (t->left == NULL&&t->right == NULL)
		{
			cout << tmp << endl;
			return tmp;
		}	
		else
		{
			return bianli(t->left, tmp) + bianli(t->right, tmp);
		}
	}
}

int main()
{
	node *T = NULL;
	T = new node(5);
	int sum=0;
	insert(T, 0);
	insert(T, 1);
	insert(T, 2);
	insert(T, 3);
	insert(T, 4);
	insert(T, 8);
	insert(T, 6);
	insert(T, 7);
	insert(T, 9);
	sum=bianli(T,0);
	//cout << "T->left:" << T->left->num << endl;
	//cout << "T->right:" << T->right->num << endl;
	cout << sum<<endl;
	system("pause");

}