此代码如何工作以找到二叉树的最小深度？

Question

I see this code from https://leetcode.com/discuss/37282/simple-python-recursive-solution-bfs-on-80ms

And it is the answer for

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.

class Solution:
        # @param {TreeNode} root
        # @return {integer}
        def minDepth(self, root):
            if not root:
                return 0

            nodes = [(root, 1)]
            while nodes:
                node, curDepth = nodes.pop(0)
                if node.left is None and node.right is None:
                    return curDepth
                if node.left:
                    nodes.append((node.left, curDepth + 1))
                if node.right:
                    nodes.append((node.right, curDepth + 1))

So my confusion is, say if node 1 has node 2 and node 3 as its .left and .right children, so the stack would be [(node 2, someDepth), (node 3 someDepth)]. Then as the stack would only pop out the last element of the list, then (node 3 someDepth) would be unpacked while node 2 is completely ignored. So in case that node 2 has no child, while node 3 has, isn't it wrong using node 3 for the subsequent iteration?

Answer 1

The point you are missing is

nodes.pop(0)

pops the 0th element.

So you are wrong here:

Then as the stack would only pop out the last element of the list, then...

Imagine a binary tree:

            1
          /    \
        2        3
     /   \     /   \
    4     5   6      7
 /   \      /   \   /
8     9    10   11 12

Here the state space will change as(for simplicity, nodes are named as their content-numbers):

# Before 1st iteration.
nodes = [(1, 1)]

# 1st iteration.
node, curDepth = 1, 1
nodes = [(2, 2), (3, 2)]

# 2nd iteration.
node, curDepth = 2, 2
nodes = [(3, 2), (4, 3), (5, 3)]

# 3rd iteration.
node, curDepth = 3, 2
nodes = [(4, 3), (5, 3), (6, 3), (7, 3)]

# 4th iteration.
node, curDepth = 4, 3
nodes = [(5, 3), (6, 3), (7, 3), (8, 4), (9, 4)]

# 5th iteration.
node, curDepth = 5, 3
# Here if node.left is None and node.right is None becomes True and curDepth i.e. 3 is returned.

As it can be seen, the nodes are processed breadth(of the tree) wise so it's a BFS.

Answer 2

The recursive solution is much easier to understand.

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

class Solution:
    # @param {TreeNode} root
    # @return {integer}
    def minDepth(self, root):
        if root is None:
            return 0

        if root.left is None:
            return 1 + self.minDepth(root.right)

        if root.right is None:
            return 1 + self.minDepth(root.left)

        return 1 + min(self.minDepth(root.left), self.minDepth(root.right))