116. Populating Next Right Pointers in Each Node

最新推荐文章于 2022-02-28 18:16:11 发布

原创最新推荐文章于 2022-02-28 18:16:11 发布 · 226 阅读

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本文介绍了一种在完美二叉树中进行层次遍历的方法，并通过使用两个队列来实现每个节点的next指针指向其右侧相邻节点的功能。在遍历过程中，通过初始化一个头节点来简化链表头部的连接。

题目描述

Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *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 toNULL.
Initially, all next pointers are set toNULL.
Note:
You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,
1
/ \
2 3
/ \ / \
4 5 6 7

After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
思路：此题相当于一个层次遍历，当每层遍历完后将next定为NULL，最开始的难题在于怎么设置链表的头节点，后来利用先初始化一个头节点head，当queue里面每出一个节点，将head的next定为该节点即可，最后上代码：

 void connect(TreeLinkNode *root) {
        if(!root) return;
        queue<TreeLinkNode*> queue1;
        queue<TreeLinkNode*> queue2;
        vector<queue<TreeLinkNode*> > v;
        v.push_back(queue1);
        v.push_back(queue2);  
        int cur=0;
        int next=1;
        v[cur].push(root);  
        while (!v[cur].empty()||!v[next].empty()) {  
            TreeLinkNode *head=new TreeLinkNode(-1); 
        while(!v[cur].empty()){
        TreeLinkNode *t = v[cur].front();  
        v[cur].pop();  
        head->next=t;
        head=head->next;
        if (t->left != NULL)  
            v[next].push(t->left);  
        if (t->right != NULL)  
            v[next].push(t->right);  
    }  
    if(v[cur].empty()){
        head->next=NULL;
        cur=1-cur;
        next=1-next;
    }
}
    }