117. Populating Next Right Pointers in Each Node II Leetcode Python

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

这题和原来题目的区别在于这里的tree有肯能不是full tree.或者有残缺。 所以最好的方法是level order travesral然后把每层的node从左边指向右边。

这里用了个额外的存储。所以空间复杂度为O(N)

一共需要遍历两遍第一遍是存储，第二遍是指针。所以时间复杂度为O(N)

code is as follow

# 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 solve(self,solution,level,root):        if len(solution)<level+1:            solution.append([])        solution[level].append(root)        if root.left:            self.solve(solution,level+1,root.left)        if root.right:            self.solve(solution,level+1,root.right)    def connect(self, root):        if root==None:            return root        solution=[]        level=0        self.solve(solution,level,root)        for level in range(len(solution)):            for count in range(1,len(solution[level])):                solution[level][count-1].next=solution[level][count]            solution[level][-1].next=None        return root                

another method need O(n) time and O(1) space

# Definition for binary tree with next pointer.# class TreeLinkNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None#         self.next = Noneclass Solution:    # @param root, a tree link node    # @return nothing    def connect(self, root):        cur=root        while cur:            pre=None            firstnode=None            while cur:                if firstnode==None:                    firstnode= cur.left if cur.left else cur.right                if cur.left:                    if pre:                        pre.next=cur.left                    pre=cur.left                if cur.right:                    if pre:                        pre.next=cur.right                    pre=cur.right                cur=cur.next            cur=firstnode