【Leetcode】111—Minimum Depth of B

一、题目描述

给定一个二叉树，找出其最小深度。
最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
说明: 叶子节点是指没有子节点的节点。
示例：

二、代码实现

方法一：BFS

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def minDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None: return 0
        queue = [root]
        layer_index = 0
        while queue:
            layer_index = layer_index + 1
            for idx in range(len(queue)):
                ele = queue.pop(0)
                if ele.left: queue.append(ele.left)
                if ele.right: queue.append(ele.right)
                if ele.left == None and ele.right == None: return layer_index

方法二：DFS

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def minDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root: return 0
        elif not root.left and root.right:
            return self.minDepth(root.right) + 1
        elif not root.right and root.left:
            return self.minDepth(root.left) + 1
        else:
            return min(self.minDepth(root.left), self.minDepth(root.right)) + 1