题目链接

Populating Next Right Pointers in Each Node II

注意点

不要访问空结点
不是完美二叉树

解法

解法一：递归，DFS。因为不是完美二叉树所以子树有可能残缺，故需要平行扫描父节点同层的节点，找到他们的左右子节点。然后右节点就是找到的节点，左节点就是如果右节点存在就选右节点否则选找到的节点

/*
// 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 NULL;
        Node* p = root->next;
        while (p) 
        {
            if (p->left) 
            {
                p = p->left;
                break;
            }
            if (p->right) 
            {
                p = p->right;
                break;
            }
            p = p->next;
        }
        if(root->right) root->right->next = p;
        if(root->left) root->left->next = root->right ? root->right : p; 
        connect(root->right);
        connect(root->left);
        return root;
    }
};

解法二：非递归。和Populating Next Right Pointers in Each Node - LeetCode一模一样

/*
// 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 NULL;
        queue<Node*> q;
        q.push(root);
        while(!q.empty())
        {
            int size = q.size();
            for(int i = 0;i < size;i++)
            {
                Node* temp = q.front();q.pop();
                if(i != size-1) temp->next = q.front();
                if(temp->left) q.push(temp->left);
                if(temp->right) q.push(temp->right);
            }
        }
        return root;
    }
};

小结

只要是遍历都有递归和非递归两种写法