671. Second Minimum Node In a Binary Tree

QUESTION:

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.

EXPLANATION:

这个思路比较简单

就是遍历之后找到第二个最小的数。看了下其他人写的

其中比较重要的信息就是 当前节点是左右两个节点中比较小的那个。

所以根节点肯定就是最小的，那么只需要寻找到第二个小的。

左右两个节点同时找最小的，再进行比较就可以。

SOLUTION:

class Solution {
    int smallest = 0;
    int secondSmallest = Integer.MAX_VALUE;
    public int findSecondMinimumValue(TreeNode root) {
        if(root==null) return -1;
        smallest = root.val;
        findSecondMinimumValueHelper(root);
        return secondSmallest==Integer.MAX_VALUE?-1:secondSmallest;
    }
    public void findSecondMinimumValueHelper(TreeNode root){
        if(root.val>smallest && root.val<secondSmallest) secondSmallest = root.val;
        if(root.left!=null) findSecondMinimumValueHelper(root.left);
        if(root.right!=null) findSecondMinimumValueHelper(root.right);
    }

}

    public int findSecondMinimumValue(TreeNode root) {

        if (root.left != null && root.right != null) {
            int left = root.left.val, right = root.right.val;
            if (root.left.val == root.val) {
                left = findSecondMinimumValue(root.left);
            }
            if (root.right.val == root.val) {
                right = findSecondMinimumValue(root.right);
            }
            if (left != -1 && right != -1) {
                return Math.min(left, right);
            }
            else return left == -1 ? right : left;
        }
        else return -1;
    }
--> return <--