二叉搜索树的第k大节点(简单)
2020年9月9日
题目来源:力扣
解题
- 暴力
中序遍历,用ArrayList记录,最后输出倒数第i个节点大小
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int kthLargest(TreeNode root, int k) {
List<Integer> list=new ArrayList<>();
find(root,list);
return list.get(list.size()-k);
}
private void find(TreeNode root,List<Integer> list){
if(root==null) return;
find(root.left,list);
list.add(root.val);
find(root.right,list);
}
}
- 逆中序遍历
左根右是从小到大,右根左是从大到小,利用全局变量k来记录是否到达想要的节点,res记录结果
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int k,res;
public int kthLargest(TreeNode root, int k) {
this.k=k;
find(root);
return res;
}
private void find(TreeNode root){
if(root == null) return;
find(root.right);
if(k == 0) return;
if(--k == 0) res = root.val;
find(root.left);
}
}