LeetCode #102 Binary Tree Level Order Traversal

Posted on | Post modified | In , , , , , , , , , ,

Problem

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

Example

Given binary tree [3,9,20,null,null,15,7],

1
2
3
4
5
  3
 / \
9  20
  /  \
 15   7

return its level order traversal as:

1
2
3
4
5
[
  [3],
  [9,20],
  [15,7]
]

Solution1

Method: Recursive Breadth First Search

Time Complexity: O(n)

Space Complexity: O(n)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def __init__(self):
        self.res=[]

    def helper(self,root,level):
        if not root:
            return None
        else:
            if level<len(self.res):
                self.res[level].append(root.val)
            else:
                self.res.append([root.val])
            self.helper(root.left,level + 1)
            self.helper(root.right,level + 1)
        return self.res
    
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        return self.helper(root,0)

Solution2

Method: Breadth First Search

Data Structure: Queue

Time Complexity: O(n)

Space Complexity: O(n)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        queue = collections.deque([(root, 0)])
        ret = []
        
        while queue:
            node, level = queue.popleft()
            if node:
                if len(ret) < level + 1:
                    ret.append([])
                ret[level].append(node.val)
                if node.left:
                    queue.append((node.left, level + 1))
                if node.right:
                    queue.append((node.right, level + 1))              

        return ret

Solution3

Method: Breadth First Search

Data Structure: Stack

Time Complexity: O(n)

Space Complexity: O(n)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []

        level = [root]
        ret = []

        while level:
            new_level = []
            ret.append([])
            for node in level:
                ret[-1].append(node.val)
                if node.left:
                    new_level.append(node.left)
                if node.right:
                    new_level.append(node.right)
            level = new_level

        return ret