python - 该代码如何找到二叉树的最小深度？

Question

我看到这段代码来自 https://leetcode.com/discuss/37282/simple-python-recursive-solution-bfs-o-n-80ms

这就是

的答案

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))

所以我的困惑是，假设节点 1 将节点 2 和节点 3 作为其 .left 和 .right 子节点，那么堆栈将是 [(node 2, someDepth), (node 3 someDepth)]。然后，由于堆栈只会弹出列表的最后一个元素，因此 (node 3 someDepth) 将被解包，而节点 2 则被完全忽略。那么，如果节点 2 没有子节点，而节点 3 有子节点，那么在后续迭代中使用节点 3 是否是错误的？

Answer 1

你错过的一点是

nodes.pop(0)

弹出第 0 个元素。

所以你错了:

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

想象一棵二叉树:

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

这里状态空间将更改为(为简单起见，节点被命名为其内容编号):

# 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.

可以看出，节点是按(树的)广度进行处理的，因此它是一个 BFS。