給定一個二叉樹

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每個 next 指針，讓這個指針指向其下一個右側節(jié)點。如果找不到下一個右側節(jié)點，則將 next 指針設置為 NULL。
初始狀態(tài)下，所有 next 指針都被設置為 NULL。

示例：

和填充每個節(jié)點的下一個右側節(jié)點指針 I的區(qū)別在于，I中的樹是完美二叉樹，node的左孩子的next必定連接node的右孩子

輸入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

輸出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

解釋：給定二叉樹如圖 A 所示，你的函數應該填充它的每個 next 指針，以指向其下一個右側節(jié)點，如圖 B 所示。

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii
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【易錯點】

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() {}

    Node(int _val, Node* _left, Node* _right, Node* _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
public:
    Node* connect(Node* root) {
        if(!root) return root;
        root->next = nullptr;
        queue<Node *> _queue;
        _queue.push(root);
        while(!_queue.empty()) {
            Node *node = _queue.front();
            _queue.pop();
            if(node->left) {
                // 右孩子存在
                if(node->right)
                    node->left->next = node->right;
                else { // 右孩子不存在
                    // 去尋找第一個存在孩子兄弟節(jié)點 【易錯點】        
                    Node *bro = node->next;
                    while(true) {
                        if(bro) {
                            if(bro->left || bro->right)
                                break;
                        } else
                            break;
                        bro = bro->next;
                    }
                    if(bro) {
                        if(bro->left)
                            node->left->next = bro->left;
                        else if(bro->right)
                            node->left->next = bro->right;
                        else
                            node->left->next = nullptr;    
                    } else
                        node->left->next = nullptr;
                }
                _queue.push(node->left);
            }
            if(node->right) {
                // 去尋找第一個存在孩子兄弟節(jié)點 【易錯點】
                Node *bro = node->next;
                while(true) {
                    if(bro) {
                        if(bro->left || bro->right)
                            break;
                    } else
                        break;
                    bro = bro->next;
                }
                if(bro) {
                    // 父節(jié)點的兄弟存在
                    if(bro->left)
                        node->right->next = bro->left;
                    else if(bro->right)
                        node->right->next = bro->right;
                    else
                        node->right->next = nullptr;
                } else
                    node->right->next = nullptr;
                _queue.push(node->right);
            }
        }
        return root;
    }
};