【Leetcode】111—Minimum Depth of B
2019-07-15 本文已影响0人
Gaoyt__
一、题目描述
给定一个二叉树,找出其最小深度。
最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
说明: 叶子节点是指没有子节点的节点。
示例:
二、代码实现
方法一:BFS
# 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 minDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root == None: return 0
queue = [root]
layer_index = 0
while queue:
layer_index = layer_index + 1
for idx in range(len(queue)):
ele = queue.pop(0)
if ele.left: queue.append(ele.left)
if ele.right: queue.append(ele.right)
if ele.left == None and ele.right == None: return layer_index
方法二:DFS
# 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 minDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root: return 0
elif not root.left and root.right:
return self.minDepth(root.right) + 1
elif not root.right and root.left:
return self.minDepth(root.left) + 1
else:
return min(self.minDepth(root.left), self.minDepth(root.right)) + 1