[leetcode] 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

思路： 对于处理不同layer的BT问题，我常用的方法就是current数组和next数组来分别储存不同layer的node, 因而这个题就非常容易解决

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None#         self.next = Noneclass Solution:    # @param root, a tree node    # @return nothing    def connect(self, root):        if root == None:            return        root.next = None        current = [root]        while len(current) != 0:            next = []            for node in current:                if node.left != None:                    next.append(node.left)                if node.right != None:                    next.append(node.right)            for i in range(0, len(next)):                if i == len(next) -1:                    next[i].next = None                else:                    next[i].next = next[i+1]            current = next