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114. 二叉树展开为链表
给你二叉树的根结点 root ，请你将它展开为一个单链表：
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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：
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树中结点数在范围
[0, 2000]内 -
-100 <= Node.val <= 100
进阶：你可以使用原地算法（\(O(1)\) 额外空间）展开这棵树吗？
思路分析
这道题本质上来讲可以说是要做一个先根遍历。但是，却可以将这个过程逆向过来，从底向上建立起关联。不可谓不精巧。
思考题：看题解中，可以逐级将左树并入到右树。尝试一下。
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一刷
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二刷
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三刷
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四刷
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五刷
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六刷
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七刷
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Flatten Binary Tree to Linked List.
* Memory Usage: 38.4 MB, less than 40.00% of Java online submissions for Flatten Binary Tree to Linked List.
*
* Copy from: https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by--26/[详细通俗的思路分析，多解法 - 二叉树展开为链表 - 力扣（LeetCode）]
*/
private TreeNode pre;
public void flatten(TreeNode root) {
if (Objects.isNull(root)) {
return;
}
flatten(root.right);
flatten(root.left);
root.right = pre;
root.left = null;
pre = root;
}
/**
* Runtime: 3 ms, faster than 29.55% of Java online submissions for Flatten Binary Tree to Linked List.
* Memory Usage: 39.8 MB, less than 10.91% of Java online submissions for Flatten Binary Tree to Linked List.
*/
public void flattenQueue(TreeNode root) {
Queue<TreeNode> queue = new ArrayDeque<>();
preorder(root, queue);
TreeNode current = queue.poll();
while (!queue.isEmpty()) {
current.right = queue.poll();
current.left = null;
current = current.right;
if (Objects.nonNull(current)) {
current.left = null;
}
}
}
private void preorder(TreeNode root, Queue<TreeNode> queue) {
if (Objects.isNull(root)) {
return;
}
queue.add(root);
preorder(root.left, queue);
preorder(root.right, queue);
}
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/**
* 参考 https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/solutions/17274/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by--26/[114. 二叉树展开为链表^] 的解法，自己实现的。
*/
public void flatten(TreeNode root) {
if (root == null) {
return;
}
while (root != null) {
if (root.left == null) {
root = root.right;
} else {
TreeNode pre = root.left;
while (pre.right != null) {
pre = pre.right;
}
pre.right = root.right;
root.right = root.left;
root.left = null;
root = root.right;
}
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-06 17:12:19
*/
public void flatten(TreeNode root) {
if (root == null) {
return;
}
while (root != null) {
if (root.left != null) {
TreeNode mostRight = root.left;
while (mostRight.right != null) {
mostRight = mostRight.right;
}
mostRight.right = root.right;
root.right = root.left;
root.left = null;
}
root = root.right;
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-06-03 20:38:46
*/
public void flatten(TreeNode root) {
if (root == null) {
return;
}
while (root != null) {
if (root.left != null) {
TreeNode right = root.left;
while (right.right != null) {
right = right.right;
}
right.right = root.right;
root.right = root.left;
root.left = null;
}
root = root.right;
}
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-11-10 22:50:45
*/
public void flatten(TreeNode root) {
if (root == null) {
return;
}
while (root != null) {
// 直接判断 root.left 是否为 null，代码更简洁
if (root.left != null) {
TreeNode mostRight = mostRight(root.left);
mostRight.right = root.right;
root.right = root.left;
root.left = null;
}
root = root.right;
}
}
private TreeNode mostRight(TreeNode root) {
TreeNode result = root;
while (result != null && result.right != null) {
result = result.right;
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-11-11 20:07:06
*/
private TreeNode head;
public void flatten(TreeNode root) {
if (root == null) {
return;
}
flatten(root.right);
flatten(root.left);
root.left = null;
root.right = head;
head = root;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2026-10-20 22:44:41
*/
public void flatten(TreeNode root) {
if (root == null) {
return;
}
if (root.left != null) {
TreeNode mostRight = root.left;
while (mostRight.right != null) {
mostRight = mostRight.right;
}
mostRight.right = root.right;
root.right = root.left;
root.left = null;
}
flatten(root.right);
}
思考题
如何使用递归来实现？
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/**
* 递归解决方案
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-06 17:32:56
*/
public void flatten(TreeNode root) {
if (root == null) {
return;
}
flatten(root.left);
flatten(root.right);
TreeNode left = root.left;
TreeNode right = root.right;
root.left = null;
root.right = left;
TreeNode mostRight = root;
while (mostRight.right != null) {
mostRight = mostRight.right;
}
mostRight.right = right;
}