116. Populating Next Right Pointers in Each Node

完美二叉树的连接

最新推荐文章于 2020-07-05 10:30:17 发布

原创最新推荐文章于 2020-07-05 10:30:17 发布 · 310 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 3 个专栏收录

67 篇文章

订阅专栏

算法面试题

50 篇文章

订阅专栏

Python

17 篇文章

订阅专栏

本文介绍了一种在完美二叉树中连接每个节点到其右侧相邻节点的方法，通过两种算法实现：广度优先搜索（BFS）和深度优先搜索（DFS）。详细解释了代码实现过程，展示了如何在Python中操作树结构。

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *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.

Example:

思路一：

与上一篇文章相同的思路，使用BFS遍历树，并在获得每一层的结点后，完成.next指针的指向

代码一：

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if root == None:
            return root
        queue = []
        queue.append(root)
        while queue:
            curtmp = []
            nextmp = []
            for node in queue:
                curtmp.append(node)
                if node.left:
                    nextmp.append(node.left)
                if node.right:
                    nextmp.append(node.right)
            queue = nextmp
            for i in range(len(curtmp)-1):
                curtmp[i].next = curtmp[i+1]
            # curtmp[-1].next = None   no necessory
        return root

思路二：

使用DFS遍历

代码二：

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left, right, next):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if root == None:
            return root
        if root.left:
            root.left.next = root.right
            if root.next:
                root.right.next = root.next.left
        self.connect(root.left)
        self.connect(root.right)
        return root

注：初始化时，所有结点的.next指针均是指向None的，因此每层末端结点没必要再做指向None的操作。