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【easy】671. Second Minimum Node In a Binary Tree

emp treenode pos turn node fin root for help

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or 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.

題目:

特殊的二叉樹,父節點是子節點中較小者,找出二叉樹中次小的值。

********這個題用java寫的…java中是null, 並且數組是‘引用’,不用 - > 用 .

Input: 
    2
   /   2   5
     /     5   7

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

Output: -1
Explanation: The smallest value is 2, but there isn‘t any second smallest value.
/**
 * 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) {
        int []data = new int []{Integer.MAX_VALUE, Integer.MAX_VALUE};
        help(root,data);
        
return data[1]!=Integer.MAX_VALUE?data[1]:-1; } public void help(TreeNode root, int []data){//java直接就是引用 if (root == null) return; if (root.val<data[0]){ data[1] = data[0]; data[0] = root.val; } if (root.val<data[1] && root.val>data[0]) data[1] = root.val; help(root.left,data); help(root.right,data); } }

【easy】671. Second Minimum Node In a Binary Tree