Leetcode 102.二叉树的层次遍历

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102.二叉树的层次遍历

题目

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

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例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
返回其层次遍历结果：

[
  [3],
  [9,20],
  [15,7]
]

方法

方法1：宽度优先遍历

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

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """

        if not root:
            return []

        queue=collections.deque([(root)])
        ans=[]
        while queue:
            level=[]
            for i in range(len(queue)):
                node=queue.popleft()
                if node:
                    level.append(node.val)
                    if node.left:
                        queue.append(node.left)
                    if node.right:
                        queue.append(node.right)
            ans.append(level)
        return ans
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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """

        if not root:
            return []

        queue=collections.deque([(root)])
        ans=[]
        while queue:
            level=[]
            q=collections.deque()
            while queue:
                node=queue.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
                level.append(node.val)
            ans.append(level)
            queue=q
        return ans
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class Solution(object):
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if root == None:
            return []
        tmp = [root]
        res = []
        while tmp:
            value = []
            l = len(tmp)
            i = 0
            while i < l:
                t = tmp.pop(0)
                value.append(t.val)
                if t.left:
                    tmp.append(t.left)
                if t.right:
                    tmp.append(t.right)
                i += 1
            res.append(value)    
        return res