[LeetCode]671. Second Minimum Node In a Binary Tree解法及python代码

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原创最新推荐文章于 2026-05-03 10:32:25 发布 359 阅读 AI 写同款· GEO 优化 ›

LeetCode --- 671. Second Minimum Node In a Binary Tree 解题报告 Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactlytwoorzerosub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes. More f... 阅读详情

原题链接：https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/
这个题的树有两个特点，一个是树如果有儿子一定有两个，另一个是根节点的值是根、左、右三个节点值最小的点。最开始的想法是基于先序遍历在树的遍历过程中剪枝，但发现这种方法在递归过程中比较难实现，就使用了另外一种办法：先序遍历整个树，在遍历过程中将值加入set中，去掉set中最小的值，如果set不为空就返回当前set中的最小值，代码如下：

class Solution(object):
    def findSecondMinimumValue(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        seMinValue = set([])
        #先序遍历
        def inOrder(node):
            if node == None:
                return
            seMinValue.add(node.val)
            if node.left != None:
                inOrder(node.left)
                inOrder(node.right)
        inOrder(root)
        if not seMinValue:
            return -1
        seMinValue.remove(min(seMinValue))
        if not seMinValue:
            return -1
        return min(seMinValue)