Second Minimum Node In a Binary Tree

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly twoor zero sub-node. If the node has two sub-nodes, then this node's value is the smaller value among its two sub-nodes.

Given such a binary tree, you need to output the second minimum value in the set made of all the nodes' value in the whole tree.

If no such second minimum value exists, output -1 instead.

Example 1:

Input: 
    2
   / \
  2   5
     / \
    5   7

Output: 5
Explanation: The smallest value is 2, the second smallest value is 5.

Example 2:

Input: 
    2
   / \
  2   2

Output: -1
Explanation: The smallest value is 2, but there isn't any second smallest value.

java 用递归写，传int 值得时候，是值传递，要注意

时间复杂度 o(n), space o(1)

这个答案已经无法通过所有test， [2,2,Integer max value],可以用更改后的code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findSecondMinimumValue(TreeNode root) {
       
       if(root == null)
           return -1;
        
        int first = root.val;
        int second = Integer.MAX_VALUE;
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while(!queue.isEmpty()){
            TreeNode node = queue.poll();
            
            
            //node.val 必须是!= first，不能是>=
            //因为如果是>=的话，第一个node时候，second会被赋予 node的值，也就是最小值，最后second返回的是最小值
            if(node.val != first && node.val < second){
                second = node.val;
            }
            
            if(node.right != null){
                queue.offer(node.right);
            }
            
            if(node.left != null)
                queue.offer(node.left);
            
        }
        
        if(second == Integer.MAX_VALUE || second == first)
            return -1;
        else
            return second;

    }
    
    
    
    
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findSecondMinimumValue(TreeNode root) {
       
       if(root == null)
           return -1;
        
        int first = root.val;
        // int second = Integer.MAX_VALUE;
            int second = -1;        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while(!queue.isEmpty()){
            TreeNode node = queue.poll();
            
            
            //node.val 必须是!= first，不能是>=
            //因为如果是>=的话，第一个node时候，second会被赋予 node的值，也就是最小值，最后second返回的是最小值
            // if(node.val != first && node.val <= second){
            //     second = node.val;
            // }
            
            if(node.val > first && (second == -1 || node.val < second)) {
                second = node.val;
            }
            
            // else if(node.val > first && node.val < second){
            //     second = node.val;
            // }
            
            if(node.right != null){
                queue.offer(node.right);
            }
            
            if(node.left != null)
                queue.offer(node.left);
            
        }
        
        if(second == -1)
            return -1;
        else
            return second;

    }
    
    
    
    
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findSecondMinimumValue(TreeNode root) {
    PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
    find(root, pq);
    if (pq.size() < 2) return -1;
    pq.poll();
    return pq.poll();
  }
  
  public void find(TreeNode node, PriorityQueue<Integer> pq) {
    if (node == null) {
      return;
    }
    if (pq.isEmpty() || node.val != pq.peek()) {
      pq.add(node.val);
    }
    
    find(node.left, pq);
    find(node.right, pq);
  }
}