[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).

1. BFS: use a queue to keep breadth-first traversal result

/**
* 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) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
// Note: The Solution object is instantiated only once and is reused by each test case.
//breadth-first traversal
queue<TreeLinkNode *> bfsq;
int levelCnt=0;
int level2=0;

if(root==NULL) return;
bfsq.push(root);
levelCnt++;
TreeLinkNode *prevList=NULL;
TreeLinkNode *topS=NULL;

while(!bfsq.empty())
{
topS=bfsq.front();
bfsq.pop();
levelCnt--;

if(topS->left!=NULL && topS->right!=NULL)
{
bfsq.push(topS->left);
level2++;
bfsq.push(topS->right);
level2++;
}

if(prevList!=NULL)
prevList->next=topS;
prevList=topS;

if(levelCnt==0)
{
levelCnt=level2;
level2=0;
prevList=NULL;
}

}
}
};

2. Recursion to save space: populating by level. node->right->next = node->next->left

class Solution {
public:
void connect(TreeLinkNode *root) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if(root==NULL || (root->left==NULL && root->right==NULL) )
return;

root->left->next=root->right;
root->right->next=(root->next)? root->next->left:NULL;

connect(root->left);
connect(root->right);
}
};