leetcood学习笔记-111-二叉树的最小深度

题目描述：

第一次提交：

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

优化后：

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

方法二：广度优先 O(N)

from collections import deque
class Solution:
    def minDepth(self, root):
        if not root:
            return 0
        else:
            node_deque = deque([(1, root),])

        while node_deque:
            depth, root = node_deque.popleft()
            children = [root.left, root.right]
            if not any(children):
                return depth
            for c in children:
                if c:
                    node_deque.append((depth + 1, c))