671. Second Minimum Node In a Binary Tree（二叉树中第二小的节点）

题目描述

给定一个非空特殊的二叉树，每个节点都是正数，并且每个节点的子节点数量只能为 2 或 0。如果一个节点有两个子节点的话，那么这个节点的值不大于它的子节点的值。

给出这样的一个二叉树，你需要输出所有节点中的第二小的值。如果第二小的值不存在的话，输出 -1 。

方法思路

Approach #1: Brute Force [Accepted]
raverse the tree with a depth-first search, and record every unique value in the tree using a Set structure uniques.

Then, we’ll look through the recorded values for the second minimum. The first minimum must be root.val\text{root.val}root.val.

class Solution{
    //Runtime: 2 ms, faster than 51.91%
    //Memory Usage: 36.9 MB, less than 19.37%
    private Set<Integer> set;
    public int findSecondMinimumValue(TreeNode root){
        set = new HashSet<>();
        dfs(root);
        int min1 = root.val;
        long ans = Long.MAX_VALUE;
        for (int v : set) {
            if (min1 < v && v < ans) ans = v;
        }
        return ans < Long.MAX_VALUE ? (int) ans : -1;
    }
    
    public void dfs(TreeNode root){
        if(root != null){
            set.add(root.val);
            dfs(root.left);
            dfs(root.right);
        }
    }
}

Approach2:
Let min1 = root.val\text{min1 = root.val}min1 = root.val. When traversing the tree at some node, node\text{node}node, if node.val > min1\text{node.val > min1}node.val > min1, we know all values in the subtree at node\text{node}node are at least node.val\text{node.val}node.val, so there cannot be a better candidate for the second minimum in this subtree. Thus, we do not need to search this subtree.

Also, as we only care about the second minimum ans\text{ans}ans, we do not need to record any values that are larger than our current candidate for the second minimum, so unlike Approach #1 we can skip maintaining a Set of values(uniques) entirely.

class Solution {
    //Runtime: 1 ms, faster than 100.00%
    //Memory Usage: 36.6 MB, less than 85.86% 
    int min1;
    long ans = Long.MAX_VALUE;

    public void dfs(TreeNode root) {
        if (root != null) {
            if (min1 < root.val && root.val < ans) {
                ans = root.val;
            } else if (min1 == root.val) {
                dfs(root.left);
                dfs(root.right);
            }
        }
    }
    public int findSecondMinimumValue(TreeNode root) {
        min1 = root.val;
        dfs(root);
        return ans < Long.MAX_VALUE ? (int) ans : -1;
    }
}