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程式語言 - LeetCode - C++ - 117. Populating Next Right Pointers in Each Node II

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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 *pre = nullptr;

            for (int i = 0; i < size; ++i) {
                Node *cur = q.front();
                q.pop();

                if (pre) {
                    pre->next = cur;
                }
                pre = cur;

                if (cur->left) {
                    q.push(cur->left);
                }
                if (cur->right) {
                    q.push(cur->right);
                }
            }
        }

        return root;
    }
};