For a given binary tree with N nodes (N>=0), find the sum of the leaf node elements.
Input format: The
first line is a non-negative integer N, which means there are N nodes
The second line is an integer k, which is the element value of the root of the tree
Next, there are N-1 lines, each line is a new node, the format is rde three integers,
r represents the element value of the parent node of the node (the parent node is guaranteed to exist); d is the direction, 0 represents the node is the left son of the parent node, 1 represents the right son; e is the element value of the node.
Output format: the
sum of leaf node elements in the tree (guaranteed to be within the range of integer variables).
Input sample:
For the binary tree in the picture:
3
20
20 0 10
20 1 25
Sample output:
35
Routine questions, save each node, when inputting, traverse the stored nodes, change the left and right nodes, pay attention to output 0 when the tree is empty, and then exit the program directly.
#include <iostream>
#include <vector>
using namespace std;
typedef struct Node {
int data;
struct Node *left;
struct Node *right;
} Node, *Tree;
int main() {
int n;
cin >> n;
if(n == 0){
cout << 0;
return 0;
}
vector<Tree> v;
Tree root = (Tree)malloc(sizeof(Node));
int k, r, d, e;
cin >> k;
root->data = k;
root->left = root->right = NULL;
v.push_back(root);
for (int i = 1; i < n; i++) {
Tree tmp = (Tree)malloc(sizeof(Node));
cin >> r >> d >> e;
tmp->data = e;
tmp->left = tmp->right = NULL;
for (int j = 0; j < v.size(); j++) {
if (v[j]->data == r) {
if (d == 0)
v[j]->left = tmp;
else
v[j]->right = tmp;
break;
}
}
v.push_back(tmp);
}
int sum = 0;
for (int i = 0; i < v.size(); i++) {
if (!v[i]->left && !v[i]->right)
sum += v[i]->data;
}
cout << sum;
return 0;
}