LeetCode 111. Minimum Depth of Binary Tree--Java, Python解法--二叉树最小高度--迭代，递归

题目地址：Minimum Depth of Binary Tree - LeetCode

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.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3/ \9  20/  \15   7

这道题目是计算二叉树最小高度，可以用迭代或者递归来做。

递归解法如下。
Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def minDepth(self, root: TreeNode) -> int:def helper(root, height):if root == None:return height-1if root.left == None and root.right == None:return heightelif root.left == None:return helper(root.right, height+1)elif root.right == None:return helper(root.left, height+1)return min(helper(root.left, height+1), helper(root.right, height+1))return helper(root, 1)

迭代的解法如下。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def minDepth(self, root: TreeNode) -> int:if root is None:return 0stack=[(1,root)]height=1while stack!=[]:depth,root=stack.pop(0)child=[root.left,root.right]if not any(child):height=max(depth,height)return heightfor i in child:if i is not None:stack.append((depth+1,i))return height

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