LeetCode111 Minimum Depth of Binary Tree

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

return its minimum depth = 2.

<思路一>我第一次用BFS做的，思路就是遍历每一层节点，每次depth+1。

如果这个节点没有左右孩子节点，将depth保存在dep列表中，最后返回dep的最小值。

如果有左孩子或右孩子，将孩子节点保存在queue中，等待下一层遍历时使用。

# 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
        """
        level = [root]
        depth = 0
        dep = []
        
        if root is None:
            return 0
        
        while level:
            queue=[]
            depth +=1
            for node in level:                     
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
                if node.left is None and node.right is None:
                    dep.append(depth)                
            level = queue
        return min(dep)

<思路二>用递归的方式，也比较简单

# 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 is None:
            return 0
        
        if root.left and root.right:
            return min(self.minDepth(root.left), self.minDepth(root.right))+1
        if root.left:
            return self.minDepth(root.left) + 1
        if root.right:
            return self.minDepth(root.right) + 1
        else:
            return 1