目录
230. 二叉搜索树中第 K 小的元素 - 力扣(LeetCode)
相当于把二叉搜索树从小到大排序,而二叉搜索树有一个特点,就是顺序左子树 < 根节点 < 右子树,因此可以考虑使用中序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
int now = 0;
int ans = 0;
public int kthSmallest(TreeNode root, int k) {
find(root, k);
return ans;
}
public void find(TreeNode root, int k) {
if (root == null) {
return;
}
find(root.left, k);
if (++now == k) {
ans = root.val;
return;
}
find(root.right, k);
}
}
199. 二叉树的右视图 - 力扣(LeetCode)
按照中 -> 右 -> 左遍历,只要每层记录一下第一个遍历到的节点
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ans = new ArrayList<>();
find(root, ans, 0);
return ans;
}
public void find(TreeNode root, List<Integer> ans, int now) {
if (root == null) {
return ;
}
if (ans.size() <= now) {
ans.add(root.val);
}
find(root.right, ans, now+1);
find(root.left, ans, now+1);
}
}
标签:right,TreeNode,val,int,HOT,100,root,LeetCode,left
From: https://www.cnblogs.com/MMMMMMMW/p/18534219