给你二叉树的根结点 root ,请你将它展开为一个单链表:
展开后的单链表应该同样使用 TreeNode ,其中 right 子指针指向链表中下一个结点,而左子指针始终为 null 。
展开后的单链表应该与二叉树 先序遍历 顺序相同。
示例 1:
输入:root = [1,2,5,3,4,null,6]
输出:[1,null,2,null,3,null,4,null,5,null,6]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [0]
输出:[0]
提示:
树中结点数在范围 [0, 2000] 内
-100 <= Node.val <= 100
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/flatten-binary-tree-to-linked-list
进阶:你可以使用原地算法(O(1)
额外空间)展开这棵树吗?
解法1:前序遍历
/**
* 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 {
//思路:1.先序遍历此二叉树,得到list,在将其转换为链表形式
public void flatten(TreeNode root) {
if(root == null){
return ;
}
List<TreeNode> list = new LinkedList<>();
preOrder(root,list);
//转换链表
for(int i = 0; i < list.size() - 1; i++){
TreeNode pre = list.get(i);
TreeNode next = list.get(i+1);
pre.right = next;
pre.left = null;
}
}
//先序遍历
public void preOrder(TreeNode tree,List<TreeNode> list){
list.add(tree);
if(tree.left != null){
preOrder(tree.left,list);
}
if(tree.right != null){
preOrder(tree.right,list);
}
}
}
解法2:尝试原地算法(O(1)
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 {
//思路:
//关键的点在于将5挪到4的右子节点
public void flatten(TreeNode root) {
if(root == null){
return ;
}
TreeNode current = root;
while(current != null){
if(current.left != null){
//当前节点的左节点不为空,若是为空,
TreeNode pre = current.left;
//找到当前子树的最右侧节点
while(pre.right != null){
pre = pre.right;
}
//将当前根节点的右子树赋值为 左子树的最右节点的右子树
pre.right = current.right;
//基于上一步的基础上,右子树已经空了,我们将左子树全部赋值到右子树,这样右子树又有值了
current.right = current.left;
//然后将左子树置为空。
current.left = null;
//轮回执行此右子树
current = current.right;
} else {
current = current.right;
}
}
}
}
解题思路:二叉树的操作,先脑海里出现它的操作步骤,再用思路不断地实现它。