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程式語言 - LeetCode - C++ - 116. Populating Next Right Pointers in Each Node
參考資訊:
https://www.cnblogs.com/grandyang/p/4288151.html
題目:

解答:
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
queue<Node*> q;
if (!root) {
return root;
}
q.push(root);
while (!q.empty()) {
int size = q.size();
Node* prev = nullptr;
for (int i = 0; i < size; ++i) {
Node* cur = q.front();
q.pop();
if (prev) {
prev->next = cur;
}
prev = cur;
if (cur->left) {
q.push(cur->left);
}
if (cur->right) {
q.push(cur->right);
}
}
}
return root;
}
};