LeetCode OJ 117. Populating Next Right Pointers in Each Node II

topic

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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
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

answer

This is a level-order traversal of a binary tree. . . AC again.

Here is the code for AC:

/**
 * 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:
    queue<TreeLinkNode *> q;
    void connect(TreeLinkNode *root) {
        if(root == NULL){
            return ;
        }
        q.push(root);
        int i;
        while(i = q.size()){
            while(i--){
                TreeLinkNode *temp = q.front();
                q.pop();
                if(i == 0){
                    temp->next = NULL;
                }
                else{
                    temp->next = q.front();
                }
                if(temp->left != NULL){
                    q.push(temp->left);
                }
                if(temp->right != NULL){
                    q.push(temp->right);
                }
            }
        }
    }
};

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