[LeetCode] 116. Populating Next Right Pointers in Each Node

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example:

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

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

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

定义Node可指向其左子树、右子树和下一个结点，下一个结点就是水平方向上的其右边的结点。当右边没有结点时，其next值为NULL。

本题给出完全二叉树，所有结点的next初始值为NULL，给各节点的next赋予正确的值。

提醒：使用固定大小的额外空间

          递归方法很好，隐式堆栈空间不算作此问题的额外空间。

要求右子树根结点的next就要有根结点的next，因此要从上到下进行操作。

方法一：使用两个迭代器，第一个迭代器沿左支从上到下遍历，另一个迭代器在此基础上从左到右对结点进行next赋值操作。

代码如下：

class Solution {
public:
    Node* connect(Node* root) {
        if(!root)return NULL;
        Node* cur=root;
        while(cur->left){
        　　Node *p=cur;
        　　while(p){
        　　　　p->left->next=p->right;
        　　　　if(p->next){
               　　p->right->next=p->next->left;
               }
               p=p->next;
           }
           cur=cur->left;
        }
    　　 return root;
    }
};

方法二：递归法。求next的值的方法和方法一相同，只是这里用递归的方法代替了从上到下的层序遍历。

代码如下：

class Solution {
public:
    Node* connect(Node* root) {
        if(!root)return NULL;
        if(root->left){
            root->left->next=root->right;
            if(root->next){
                root->right->next=root->next->left;
            }
        }
        connect(root->left);
        connect(root->right);
        return root;
    }
};

方法三：使用堆栈代替递归操作。将每一层的结点从左向右依次放如queue中，然后从头依次取出结点，同时将下一层的结点放进去。

代码如下：

class Solution {
public:
    Node* connect(Node* root) {
        if(!root)return NULL;
        queue<Node*> q;
        q.push(root);
        while(!q.empty()){
            int len=q.size();
            while(len){
                Node* cur=q.front();
                q.pop();
                if(--len)cur->next=q.front();
                if(cur->left)q.push(cur->left);
                if(cur->right)q.push(cur->right);
            }
        }
        return root;
    }
};