[leetcode]Populating Next Right Pointers in Each Node II @ Python

原题地址:https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

题意:

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

  • You may only use constant extra space.

 

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

 

After calling your function, the tree should look like:

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

解题思路:和"Populating Next Right Pointers in Each Node"这道题不同的一点是,这道题的二叉树不是满的二叉树,有些节点是没有的。但是也可以按照递归的思路来完成。在编写递归的基准情况时需要将细节都考虑清楚:

代码一:

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if root:
            if root.left and root.right:
                root.left.next = root.right
                tmp = root.next
                while tmp:
                    if tmp.left: root.right.next = tmp.left; break
                    if tmp.right: root.right.next = tmp.right; break
                    tmp = tmp.next
            elif root.left:
                tmp = root.next
                while tmp:
                    if tmp.left: root.left.next = tmp.left; break
                    if tmp.right: root.left.next = tmp.right; break
                    tmp = tmp.next
            elif root.right:
                tmp = root.next
                while tmp:
                    if tmp.left: root.right.next = tmp.left; break
                    if tmp.right: root.right.next = tmp.right; break
                    tmp = tmp.next
            self.connect(root.right)
            self.connect(root.left)
            # @connect(root.right)should be the first!!!
            

代码二:

思路更加精巧,代码更加简洁。

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree node
    # @return nothing
    def connect(self, root):
        if root:
            p = root; q = None; nextNode = None
            while p:
                if p.left:
                    if q: q.next = p.left
                    q = p.left
                    if nextNode == None: nextNode = q
                if p.right:
                    if q: q.next = p.right
                    q = p.right
                    if nextNode == None: nextNode = q
                p = p.next
            self.connect(nextNode)

 

 

posted @ 2014-05-22 12:20  南郭子綦  阅读(1828)  评论(0编辑  收藏  举报