力扣labuladong——一刷day36

提示：文章写完后，目录可以自动生成，如何生成可参考右边的帮助文档

---

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

/**
 * 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

/**
 * 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

/**
 * 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

/**
 * 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);
    }
}