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题目:
Given a binary tree, find its minimum depth.The minimum depth is the
number of nodes along the shortest path from the root node down to the
nearest leaf node.
解析:
找二叉树的根节点离最近叶节点的深度
1,判断当前结点的左右子树是否为空
2,若左右子树均为空,则返回值1
3,若左子树为空,右子树不为空,则返回当前结点的右子结点的深度值+1
4,若右子树为空,左子树不为空,则返回当前结点的左子结点的深度值+1
5,若左右子树均不为空,则返回当前结点的左、右子结点的深度值的更小的那个值+1
代码:
def mindepth(root):
tmp = root
if(tmp.left == None and tmp.right == None):
return 1
elif(tmp.left == None):
return mindepth(tmp.right) + 1
elif(tmp.right == None):
return mindepth(tmp.left) + 1
else:
return(1+min(mindepth(tmp.left),mindepth(tmp.right)))
全部代码:
class treenode :
def __init__(self,val = None,left = None,right = None):
self.val = val
self.left = left
self.right = right
def settag(self,tag = None):
self.tag = tag
def insertnode(root,treenode):
tmp = root
if (root == None):
root = treenode
while(tmp != None):
if(treenode.val == tmp.val):
break
elif(treenode.val <= tmp.val):
if(tmp.left != None):
tmp = tmp.left
else:
tmp.left = treenode
else:
if(tmp.right != None):
tmp = tmp.right
else:
tmp.right = treenode
return root
def mindepth(root):
tmp = root
if(tmp.left == None and tmp.right == None):
return 1
elif(tmp.left == None):
return mindepth(tmp.right) + 1
elif(tmp.right == None):
return mindepth(tmp.left) + 1
else:
return(1+min(mindepth(tmp.left),mindepth(tmp.right)))
if __name__=='__main__':
t=treenode()
L = [8,3,9,2,12,5,34,23,10]
t.val = L[0]
for x in L:
node = treenode()
node.val = x
t = insertnode(t,node)
print(mindepth(t))
print(maxdepth(t))