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

1 /**
2  * Definition for binary tree with next pointer.
3  * struct TreeLinkNode {
4  *  int val;
5  *  TreeLinkNode *left, *right, *next;
6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
7  * };
8  */
9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         if (!root || !root->left) return;
13
14         root->left->next = root->right;
15         if (root->next)
16             root->right->next = root->next->left;
17
18         connect(root->left);
19         connect(root->right);
20     }
21 };

1 /**
2  * Definition for binary tree with next pointer.
3  * struct TreeLinkNode {
4  *  int val;
5  *  TreeLinkNode *left, *right, *next;
6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
7  * };
8  */
9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         if (!root || !root->left) return;
13
14         queue<TreeLinkNode* > qu;
15         qu.push(root);
16         while (!qu.empty()) {
17             int n = qu.size();
18             while (n--) {
19                 TreeLinkNode* temp = qu.front();
20                 qu.pop();
21
22                 if (temp->left) {
23                     qu.push(temp->left);
24                     qu.push(temp->right);
25                     temp->left->next = temp->right;
26                     if (temp->next)
27                         temp->right->next = temp->next->left;
28                 } else {
29                     return;
30                 }
31             }
32         }
33     }
34 };