【leetcode】 Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

benjamin_sunny_li

最新推荐文章于 2021-09-24 12:05:40 发布

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本文链接：https://blog.youkuaiyun.com/weixin_38111819/article/details/80156326

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本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使其指向其右侧相邻节点的算法实现。该方法采用层序遍历的方式，并且仅使用常数额外空间。

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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.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

After calling your function, the tree should look like:

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

思路：还是层序遍历的变形（又叫二叉树BFS）

# 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 not root:
            return 
        queue = [root]
        last = root
        while queue:
            node = queue.pop(0)
            if node.left:
                queue.append(node.left)
            if node.right:
                queue.append(node.right)
            if node == last:
                node.next = None
                if queue:
                    last = queue[-1]
            else:
                node.next = queue[0]