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文章目录
- 前言
- [一、力扣230. 二叉搜索树中第K小的元素](#一、力扣230. 二叉搜索树中第K小的元素)
- [二、力扣538. 把二叉搜索树转换为累加树](#二、力扣538. 把二叉搜索树转换为累加树)
- [三、力扣1038. 从二叉搜索树到更大和树](#三、力扣1038. 从二叉搜索树到更大和树)
前言
首先,BST 的特性大家应该都很熟悉了: 1、对于 BST 的每一个节点 node,左子树节点的值都比 node 的值要小,右子树节点的值都比 node 的值大。 2、对于 BST 的每一个节点 node,它的左侧子树和右侧子树都是 BST。
一、力扣230. 二叉搜索树中第K小的元素
java
/**
* 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 {
List<Integer> list = new ArrayList<>();
public int kthSmallest(TreeNode root, int k) {
fun(root);
return list.get(k-1);
}
public void fun(TreeNode root){
if(root == null){
return ;
}
fun(root.left);
list.add(root.val);
fun(root.right);
}
}
不使用额外空间
java
/**
* 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 res = 0,count = 0;
public int kthSmallest(TreeNode root, int k) {
fun(root,k);
return res;
}
public void fun(TreeNode root,int k){
if(root == null){
return ;
}
fun(root.left,k);
count ++;
if(count == k){
res = root.val;
return ;
}
fun(root.right,k);
}
}
二、力扣538. 把二叉搜索树转换为累加树
java
/**
* 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 count = 0;
public TreeNode convertBST(TreeNode root) {
fun(root);
return root;
}
public void fun(TreeNode root){
if(root == null){
return ;
}
fun(root.right);
count += root.val;
root.val = count;
fun(root.left);
}
}
三、力扣1038. 从二叉搜索树到更大和树
java
/**
* 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 count = 0;
public TreeNode bstToGst(TreeNode root) {
fun(root);
return root;
}
public void fun(TreeNode root){
if(root == null){
return;
}
fun(root.right);
count += root.val;
root.val = count;
fun(root.left);
}
}