題目 Binary Tree Level Order Traversal

層次遍歷二叉樹
鏈接
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

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

3
/ \
9 20
/ \
15 7

return its level order traversal as:

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

解答

my answer

class Solution:def levelOrder(self, root):""":type root: TreeNode:rtype: List[List[int]]"""res, level = [], [root]while root and level:currentNode = []nextLevel = []for node in level:currentNode.append(node.val)if node.left:nextLevel.append(node.left)if node.right:nextLevel.append(node.right)res.append(currentNode)level = nextLevelreturn res

python 解法2

class Solution:# 返回從上到下每個節(jié)點(diǎn)值列表，例：[1,2,3]def PrintFromTopToBottom(self, root):# write code hereif root is None:return []queue = [root]res = []while queue:node = queue.pop(0)res.append(node.val)if node.left:queue.append(node.left)if node.right:queue.append(node.right)return res

網(wǎng)友值得參考的解答

Solution 1, (6 lines)

def levelOrder(self, root):ans, level = [], [root]while root and level:ans.append([node.val for node in level])LRpair = [(node.left, node.right) for node in level]level = [leaf for LR in LRpair for leaf in LR if leaf]return ans

Solution 2, (5 lines), same idea but use only one list comprehension in while loop to get the next level

def levelOrder(self, root):ans, level = [], [root]while root and level:ans.append([node.val for node in level])            level = [kid for n in level for kid in (n.left, n.right) if kid]return ans

Solution 3 (10 lines), just an expansion of solution 1&2 for better understanding.

def levelOrder(self, root):if not root:return []ans, level = [], [root]while level:ans.append([node.val for node in level])temp = []for node in level:temp.extend([node.left, node.right])level = [leaf for leaf in temp if leaf]return ans

other

def levelOrder(self, root):ret = []  level = [root]while root and level:currentNodes = []nextLevel = []for node in level:currentNodes.append(node.val)if node.left:nextLevel.append(node.left)if node.right:nextLevel.append(node.right)ret.append(currentNodes)level = nextLevelreturn ret

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