LeetCode Q117: Populating Next Right Pointers in Each Node II (hard)

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

Solution:

This question is not hard at all and can be solved by my preivous solutoin. However, since in my previous solution, I used at most O(n) space to save intermediate results which violate the requirement of the question, I get rid of using array to store intermediate results in this question. Instead, all intermediate results can be tracked using a pointer.

	/**
	* 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) {
	queue<TreeLinkNode*> Q;
	if(root==NULL)
	return;
	Q.push(root);
	int cur=1;
	int next=0;
	TreeLinkNode* previous=NULL;
	while(!Q.empty()){
	TreeLinkNode* front=Q.front();
	Q.pop();
	cur--;
	if(front->left!=NULL){
	Q.push(front->left);
	next++;
	}
	if(front->right!=NULL){
	Q.push(front->right);
	next++;
	}
	if(previous!=NULL){
	previous->next=front;
	previous=front;
	}else{
	previous=front;
	}
	if(cur==0){
	previous->next=NULL;
	previous=NULL;
	cur=next;
	next=0;
	}
	}
	}
	};

view raw Q117.cpp hosted with ❤ by GitHub