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algorithm notes: leetcode#671 second minimum node in a binary tree

admin 11月 12, 2020 0

Problem

Solution

Initial thoughts

Python implementation

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def findSecondMinimumValue(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self._min_val = float("inf")
        self._sec_min = float("inf")
        def dfs(node):
            if node is None:
                return node
            if node.val < self._min_val:
                self._min_val, self._sec_min = node.val, self._min_val
            if self._min_val < node.val < self._sec_min:
                self._sec_min = node.val
            dfs(node.left)
            dfs(node.right)
        
        dfs(root)
        return self._sec_min if self._sec_min < float("inf") else -1

Java implementation

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    private int minVal = Integer.MAX_VALUE;
    private int secVal = Integer.MAX_VALUE;
    private void dfs(TreeNode node){
        if(node==null){ return ; }
        if(node.val < minVal){
            secVal = minVal;
            minVal = node.val;
        }
        if(minVal < node.val && node.val < secVal){
            secVal = node.val;
        }
        dfs(node.left);
        dfs(node.right);
    }
    public int findSecondMinimumValue(TreeNode root) {
        dfs(root);
        return secVal < Integer.MAX_VALUE ? secVal : -1;
    }
}

Time complexity

O(n)

Space complexity

O(n)

Links

671. Second Minimum Node In a Binary Tree
(中文版) 算法笔记: 力扣#671 二叉树中第二小的节点