描述
给定一个二叉树,找出其最小深度。
二叉树的最小深度为根节点到最近叶子节点的距离。
您在真实的面试中是否遇到过这个题?
是
样例
给出一棵如下的二叉树:
1
/ \
2 3
/ \
4 5
这个二叉树的最小深度为 2
解题思路:
利用了二叉树的层次遍历的思想,在每一层的遍历中都判断一下是否有叶子结点,如果有叶子结点,则返回该层深度。
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of binary tree
@return: An integer
"""
def minDepth(self, root):
# write your code here
self.depth = 0
if not root:
return self.depth
q = [root]
while q:
new_q = []
self.depth += 1
for node in q:
if node.left:
new_q.append(node.left)
if node.right:
new_q.append(node.right)
# 判断节点是否为叶子节点, 如果出现叶子节点,即可返回最小深度
if node.left == None and node.right == None:
return self.depth
q = new_q
return self.depth
"""
@param root: The root of binary tree
@return: An integer
"""
def minDepth(self, root):
# write your code here
self.depth = 0
if not root:
return self.depth
q = [root]
while q:
new_q = []
self.depth += 1
for node in q:
if node.left:
new_q.append(node.left)
if node.right:
new_q.append(node.right)
# 判断节点是否为叶子节点, 如果出现叶子节点,即可返回最小深度
if node.left == None and node.right == None:
return self.depth
q = new_q
return self.depth