moonlightshadow123

6/22/2017 - 7:23 AM

- Populating Next Right Pointers in Each Node

```
# Definition for binary tree with next pointer.
# class TreeLinkNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# self.next = None
class Solution:
# @param root, a tree link node
# @return nothing
def connect(self, root):
if root == None:
return
queue = [root]
while len(queue) != 0:
levelNum = len(queue)
prev = None
for i in range(levelNum):
curNode = queue[0]
del queue[0]
curNode.next = prev
prev = curNode
# Noet that we first append the right nodes, then the left nodes,
# so that you can do `curNode.next = prev`
if curNode.right:
queue.append(curNode.right)
if curNode.left:
queue.append(curNode.left)
```

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/#/description

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
```