题目来源:
leetcode题目,网址:LCR 052. 递增顺序搜索树 - 力扣(LeetCode)
解题思路:
中序遍历时修改指针即可。
解题代码:
- /**
- * 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) {
- TreeNode head=new TreeNode();//空头
- transfer(root,head);
- return head.right;
- }
- public TreeNode transfer(TreeNode root,TreeNode pre){
- if(root.left!=null){
- pre=transfer(root.left,pre);
- }
- root.left=null;
- pre.right=root;
- pre=pre.right;
- if(root.right!=null){
- pre=transfer(root.right,pre);
- }
- pre.right=null;
- return pre;
- }
- }
总结:
官方题解给出了两种解法。第一种是中序遍历记录值,然后生成一颗新的二叉树。第二种是中序遍历是修改节点指向。