[数据结构与算法]671. 二叉树中第二小的节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    Set<Integer> set = new HashSet<>();
    public int findSecondMinimumValue(TreeNode root) {
        
        dfs(root);
        int fir = Integer.MAX_VALUE, sec = Integer.MAX_VALUE;
        if (set.size() < 2) {
            return -1;
        }
        for (Integer i : set) {
            if (i <= fir) {
                sec = fir;
                fir = i;
            } else if (i <= sec){
                sec = i;
            }
        }
        
        return sec;
    }
    void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        set.add(root.val);
        dfs(root.left);
        dfs(root.right);
    }
}

总结：首先采用递归将tree的val值存入到set的不重复的集合中，然后用两个变量得出，第一大和第二大，最后经过判断可得出最后结论