Leetcode|Populating Next Right Pointers in Each Node

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 to NULL.

Initially, all next pointers are set to NULL.

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

解法1：BFS搜索二叉树，把所有节点放入数组中，利用完全二叉树的性质。可以知道那个节点的next为NULL。

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  TreeLinkNode *left, *right, *next; *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */void connect(TreeLinkNode *root) {        if(!root) return;        queue<TreeLinkNode *> q;        q.push(root);        vector<TreeLinkNode *> res;        while(q.size()>0){            TreeLinkNode* p=q.front();            q.pop();            res.push_back(p);            if(p->left) q.push(p->left),q.push(p->right);        }        int count=2;        for(int i=0;i<res.size();i++){            if(i!=count-2) res[i]->next=res[i+1];            else {                count<<=1;            }        }    }

解法2：递归，利用一个关键信息，就是如果root有next，那么就知道他的堂兄弟的节点了。

void connect(TreeLinkNode *root){//没有定义的->next就是NULL了。和struct的定义有关系的。这个很重要。24ms    if(root==NULL) return;    if(root->left&&root->right){        root->left->next=root->right;    }    if(root->next&&root->right){        root->right->next=root->next->left;    }    connect(root->left);    connect(root->right);}