题解 | #二叉树的最小深度#

思路：使用层次遍历的思想 当 节点左右孩子都是空返回当前层

#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def run(self , root: TreeNode) -> int:
        if root is None:return 0
        stack=[root]
        ceng=0
        while stack:
            for i in range(len(stack)):
                node=stack.pop(0)
                if node.left:stack.append(node.left)
                if node.right:stack.append(node.right)
                if node.left is None and node.right is None:return ceng+1
            ceng+=1
        return ceng+1