LeetCode | 0513. 找树左下角的值【Python】

Wonz 2021-01-20 23:11:42
Python github def


Problem

LeetCode

Given the root of a binary tree, return the leftmost value in the last row of the tree.

Example 1:

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Input: root = [2,1,3]
Output: 1

Example 2:

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Input: root = [1,2,3,4,null,5,6,null,null,7]
Output: 7

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -231 <= Node.val <= 231 - 1

问题

力扣

给定一个二叉树,在树的最后一行找到最左边的值。

示例 1:

输入:
2
/ \
1 3
输出:
1

示例 2:

输入:
1
/ \
2 3
/ / \
4 5 6
/
7
输出:
7

注意: 您可以假设树(即给定的根节点)不为 NULL

思路

BFS

常规 BFS 是先上后下,先左后右层次遍历。我们可以改变一下 BFS 遍历顺序,先上后下,先右后左,这样最后遍历的一个节点一定是左下角节点,最后直接返回节点值就行。

Python3 代码

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def findBottomLeftValue(self, root: TreeNode) -> int:
import collections
q = collections.deque()
q.append(root)
while q:
node = q.popleft()
 # 先右后左
if node.right:
q.append(node.right)
if node.left:
q.append(node.left)
 # 最后一个必是最左下角的节点
return node.val

GitHub 链接

Python

版权声明
本文为[Wonz]所创,转载请带上原文链接,感谢
https://my.oschina.net/wonz/blog/4916952

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