LeetCode 剑指 Offer II 052. 展平二叉搜索树
文章目录
题目描述
LeetCode 剑指 Offer II 052. 展平二叉搜索树
提示:
树中节点数的取值范围是 [1, 100]
0 <= Node.val <= 1000
一、解题关键词
二、解题报告
1.思路分析
2.时间复杂度
3.代码示例
方法一
/**
* 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 TreeNode increasingBST(TreeNode root) {
//递归展开
List<Integer> res = new ArrayList<>();
inorder(root,res);
TreeNode dummyNode = new TreeNode(-1);
TreeNode currNode = dummyNode;
for(int value : res){
currNode.right = new TreeNode(value);
currNode = currNode.right;
}
return dummyNode.right;
}
//中序遍历
void inorder(TreeNode root,List<Integer> res){
if(root == null) return;
inorder(root.left,res);
res.add(root.val);
inorder(root.right,res);
}
}
方法二
/**
* 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 TreeNode increasingBST(TreeNode root) {
//递归展开
List<Integer> res = new ArrayList<>();
inorder(root,res);
TreeNode dummyNode = new TreeNode(-1);
TreeNode currNode = dummyNode;
for(int value : res){
currNode.right = new TreeNode(value);
currNode = currNode.right;
}
return dummyNode.right;
}
//中序遍历
void inorder(TreeNode root,List<Integer> res){
if(root == null) return;
inorder(root.left,res);
res.add(root.val);
inorder(root.right,res);
}
}
2.知识点
树的遍历 遍历之后重新组装数据
总结
相同题目
897. 递增顺序搜索树