题：
给定一个二叉树根节点，获取从根到叶节点形成的所有节点的值之和。
解：
从上面的图像中，我们有4个叶节点，即：3、4、6、7
所以路径将是
1-> 2 -> 3 Number = 123
1-> 2 -> 4 Number = 124
1-> 5 -> 6 Number = 156
1-> 5 -> 7 Number = 157
因此总和为：
123 + 124 + 156 + 157 = 560
解决方案是进行预遍历并跟踪计算到该节点的值。
对于每个节点，将值乘以10 +节点’s data.
C ++解决方案
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
// structure to hold binary tree node
struct Node
{
int data;
Node *left, *right;
Node(int data)
{
this->data = data;
this->left = this->right = nullptr;
}
};
int value_root_to_leaf;
void sum_root_to_leaf(Node *node, int i)
{
if (node == NULL)
{
return;
}
if (node->left == NULL && node->right == NULL)
{
value_root_to_leaf = value_root_to_leaf + (i * 10 + node->data);
return;
}
sum_root_to_leaf(node->left, i * 10 + node->data);
sum_root_to_leaf(node->right, i * 10 + node->data);
}
int main()
{
Node* root = nullptr;
/* Binary tree:
1
/ \
/ \
2 5
/ \ / \
/ \ / \
3 4 6 7
*/
root = new Node(1);
root->left = new Node(2);
root->right = new Node(5);
root->left->left = new Node(3);
root->left->right = new Node(4);
root->right->left = new Node(6);
root->right->right = new Node(7);
sum_root_to_leaf(root, 0);
cout<<"The sum of root to leaf is "<<value_root_to_leaf<<endl;
return 0;
}
输出量
The sum of root to leaf is 560
