每日一道Leetcode - 671. 二叉树中第二小的节点

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def findSecondMinimumValue(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # 最小的一定是根节点
        # 节点值大于根节点的子树不搜索
        if root is None: return -1;
        
        def help(node, min_data):
            if node is None: return -1;
            if(node.val>min_data): return node.val;
            left = help(node.left,min_data)
            right = help(node.right,min_data)
            # 左空情况 右空情况
            if(left == -1): return right;
            if(right==-1): return left;

            return min(left,right);

        return help(root,root.val)