[LeetCode] 117. Populating Next Right Pointers in Each Node II

内容导读

互联网集市收集整理的这篇技术教程文章主要介绍了[LeetCode] 117. Populating Next Right Pointers in Each Node II，小编现在分享给大家，供广大互联网技能从业者学习和参考。文章包含2587字，纯文字阅读大概需要4分钟。

内容图文

Given a binary tree

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

Explanation: Given the above 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.

题目大意：

 

方法：

方法一：迭代法

看到图片的第一想法是使用队列进行层序遍历。

本题和116题很类似，只不过本题中的二叉树不再是完全二叉树。方法和116一模一样。

116题参考：https://www.cnblogs.com/cff2121/p/11008784.html

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 size=q.size();
            for(int i=0;i<size;++i){
                Node* cur=q.front();
                q.pop();
                if(i<size-1)cur->next=q.front();
                
                if(cur->left)q.push(cur->left);
                if(cur->right)q.push(cur->right);
            }
        }
        return root;
    }
};

 

方法二：递归

递归层序遍历，然后给每个元素的next指定节点。

代码如下：

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;
    }
};

 

方法三：使用迭代器

依旧是使用两个迭代器，一个用于从上向下移动，一个用于每层的从左向右移动。

代码如下：

class Solution {
public:
    Node* connect(Node* root) {
        if(!root)return NULL;
        Node *dummy=new Node(0,NULL,NULL,NULL),*cur=dummy,*temp=root;
        while(temp){
            if(temp->left){
                cur->next=temp->left;
                cur=cur->next;
            }
            if(temp->right){
                cur->next=temp->right;
                cur=cur->next;
            }
            temp=temp->next;
            if(!temp){
                cur=dummy;
                temp=dummy->next;
                dummy->next=NULL;
            }
        }
        return root;
    }
};

 

内容总结

以上是互联网集市为您收集整理的[LeetCode] 117. Populating Next Right Pointers in Each Node II全部内容，希望文章能够帮你解决[LeetCode] 117. Populating Next Right Pointers in Each Node II所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错，欢迎将互联网集市网站推荐给程序员好友。

内容备注

版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 gblab@vip.qq.com 举报，一经查实，本站将立刻删除。

内容手机端

---