题解 | #二叉搜索树的第k个节点#
二叉搜索树的第k个节点
http://www.nowcoder.com/practice/57aa0bab91884a10b5136ca2c087f8ff
采用递归的方式中根遍历
先从小到大排序放进list中,再根据索引get
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* public TreeNode(int val) {
* this.val = val;
* }
* }
*/
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param proot TreeNode类
* @param k int整型
* @return int整型
*/
public int KthNode (TreeNode proot, int k) {
// write code here
if(proot==null) return -1;
// 先找到最小值
LinkedList<TreeNode> linkedlist = new LinkedList<>();
LinkedList<Integer> list =inorder(proot,new LinkedList<>());
return k>0&&k<=list.size()?list.get(k-1):-1;
}
private LinkedList<Integer> inorder(TreeNode treenode,LinkedList<Integer> linkedlist){
if(treenode!=null){
inorder(treenode.left,linkedlist);
linkedlist.add(treenode.val);
inorder(treenode.right,linkedlist);
return linkedlist;
}
return linkedlist;
}
}