[LeetCode] 111. Minimum Depth of Binary Tree_Easy tag:DFS

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.

 2018/0813 update: 不能直接用l or r, 因为我们要先选小的, 如果小的是空, 然后才需要去判断那个大的. 所以需要min 及max, 而且min应该在前面.

思路为DFS recursive, 但是需要注意的是, 比如[1,2] 应该return 2, 但是上面知会return 1. 所以需要return min(l,r) or max(l, r), 因为有一边有, 另一边没有的时候要返回有的. 

 2020/0509 update: 直接用BFS，碰到第一个leaf，return当时的height即可。

1. Constraints

1) can be none => 0

 

2. Ideas

BFS T: O(n)   S: O(n)

DFS recursive    T: O(n)   S: O(n)

 

3, Code

class Solution: 
    def minDepth(self, root):
        if not root: return 0
        l, r = self.minDepth(root.left), self.minDepth(root.right)
        return 1+ (min(l, r) or max(l, r))

 

=> more elegant

class Solution:
    def minDepth(self, root):
        if not root: return 0
        d = map(self.minDepth, (root.left, root.right))
        return 1 + (min(d) or max(d))

 

利用BFS

class Solution:
    def miniDepth(self, root):
        if not root: return 0
        queue = collections.deque([(root, 1)])
        while queue:
            node, height = queue.popleft()
            if not node.left and not node.right:
                return height
            if node.left:
                queue.append((node.left, height + 1))
            if node.right:
                queue.append((node.right, height + 1))
    

 

4. Test cases

1) [1,2]

2) [3,9,20,null,null,15,7]