（python) leetcode刷题——Minimum Depth of Binary Tree

最新推荐文章于 2025-08-16 00:47:01 发布

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本文介绍了一种求解二叉树最小深度的算法，通过递归方式找到从根节点到最近叶节点的路径长度。文章详细解析了算法的实现步骤，并提供了完整的Python代码示例。

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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))