# LeetCode 116:Populating Next Right Pointers in Each Node

Datetime:2016-08-23 01:23:07         Topic: LeetCode          Share        Original >>

Given a binary tree

```struct TreeLinkNode {
}```

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```

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```//对二叉树进行先序遍历，根据题述，给定的是满二叉树，左孩子为空，则右孩子一定为空，所以左孩子为空，则return
//如果左孩子不为空，则右孩子一定不为空，所以链接左孩子和右孩子即可（左孩子的next赋值为右孩子）;
//由于先序遍历，所以父节点的next比子节点的next先被设置，故父节点不同的两个子节点进行连接，就可以用到父节点的next，
//即root->right->next = root->next->left;
class Solution {
public:
if (root == NULL || root->left==NULL)
return;

if (root->left != NULL)
root->left->next = root->right;

if (root->right != NULL && root->next != NULL)
root->right->next = root->next->left;
else if (root->right != NULL)
root->right->next = NULL;

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