Populating Next Right Pointers in Each Node II(BFS)

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 toNULL.

Initially, all next pointers are set toNULL.

Note:

• You may only use constant extra space.

• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

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

分析

BFS，最后一个node记得用i== n-1判断

# 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):        if not root:            return root        q=[root]        while q:            n = len(q)            for i in range(n):                cur = q.pop(0)                if i == n-1:                    cur.next =None                else:                    cur.next = q[0]                if cur.left:                    q.append(cur.left)                if cur.right:                    q.append(cur.right)

利用链表特性。每层都建个dummy。然后dummy指向前一层root。左右依次加入链表

class Solution:   
	def connect(self, root):        
		while root:
			dummy = TreeLinkNode(0)
			curc = dummy
			while root:根据本层的遍历连接孩子层
					if root.left:
			        curc.next = root.left
			        curc = curc.next
			    if root.right:
			        curc.next = root.right
			        curc = curc.next
			    root = root.next
			root = dummy.next	#root设置为孩子层	

DFS Recursive

"""
# 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 not root:
            return None
        p = pre = Node(-1,None,None,None)
        res = root
        while root:
            if root.left:
                pre.next = root.left
                pre = pre.next
            if root.right:
                pre.next = root.right
                pre = pre.next
            root = root.next
        self.connect(p.next)
        return res