题目LeetCode114
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。
输入： root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]
Python解法
解法一（前序遍历）
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
preorderList = list()
def preorder(root: TreeNode):
if root:
preorderList.append(root)
preorder(root.left)
preorder(root.right)
preorder(root)
size = len(preorderList)
for i in range(1, size):
pre, cur = preorderList[i - 1], preorderList[i]
pre.left = None
pre.right = cur解法二（遍历并生成）
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
if not root:
return
stack = [root]
pre = None
while stack:
cur = stack.pop()
if pre:
pre.left = None
pre.right = cur
left, right = cur.left, cur.right
if right:
stack.append(right)
if left:
stack.append(left)
pre = cur过程演示
解法三（前驱结点）
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def flatten(self, root: Optional[TreeNode]) -> None:
"""
Do not return anything, modify root in-place instead.
"""
cur = root
while cur:
if cur.left:
predecessor = cur.left
while predecessor:
predecessor = precedecessor.right
predecessor.right = cur.right
cur.right = cur.left
cur.left = None
cur = cur.right过程演示
Java解法
解法一（前序遍历）
java
/**
* 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 {
// 存储前序遍历所有节点
private List<TreeNode> preList = new ArrayList<>();
public void flatten(TreeNode root) {
// 1. 递归前序遍历收集节点
preOrderTraverse(root);
int size = preList.size();
// 2. 遍历列表，拼接成单链表
for (int i = 1; i < size; i++) {
TreeNode prev = preList.get(i - 1);
TreeNode curr = preList.get(i);
prev.left = null; // 左指针置空
prev.right = curr; // 右指针指向下一节点
}
}
// 递归前序遍历：根 -> 左 -> 右
private void preOrderTraverse(TreeNode root) {
if (root == null) return;
preList.add(root);
preOrderTraverse(root.left);
preOrderTraverse(root.right);
}
}解法二（遍历并生成）
java
/**
* 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 void flatten(TreeNode root) {
if (root == null) return;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
TreeNode prev = null; // 记录上一个处理的节点
while (!stack.isEmpty()) {
TreeNode curr = stack.pop();
// 上一节点指向当前节点
if (prev != null) {
prev.left = null;
prev.right = curr;
}
// 先存左右子节点，栈后进先出，先压右再压左保证前序
TreeNode left = curr.left;
TreeNode right = curr.right;
if (right != null) {
stack.push(right);
}
if (left != null) {
stack.push(left);
}
// 更新前驱节点
prev = curr;
}
}
}解法三（前驱结点）
java
/**
* 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 void flatten(TreeNode root) {
TreeNode curr = root;
// 逐个向右遍历节点
while (curr != null) {
// 当前节点存在左子树才需要调整
if (curr.left != null) {
// nxt 保存左子树头部
TreeNode nxt = curr.left;
TreeNode predecessor = nxt;
// 找到左子树最右侧节点
while (predecessor.right != null) {
predecessor = predecessor.right;
}
// 左子树末尾接上原右子树
predecessor.right = curr.right;
// 左子树移到右侧，清空左指针
curr.left = null;
curr.right = nxt;
}
// 处理下一个节点
curr = curr.right;
}
}
}C++解法
解法一（前序遍历）
cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
vector<TreeNode*> preList;
// 前序递归收集节点
void preOrderTraverse(TreeNode* root) {
if (!root) return;
preList.push_back(root);
preOrderTraverse(root->left);
preOrderTraverse(root->right);
}
public:
void flatten(TreeNode* root) {
preOrderTraverse(root);
int size = preList.size();
// 节点串联成右斜单链表
for (int i = 1; i < size; ++i) {
TreeNode* prev = preList[i - 1];
TreeNode* curr = preList[i];
prev->left = nullptr;
prev->right = curr;
}
}
};解法二（遍历并生成）
cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void flatten(TreeNode* root) {
if (!root) return;
stack<TreeNode*> stk;
stk.push(root);
TreeNode* prev = nullptr;
while (!stk.empty()) {
TreeNode* curr = stk.top();
stk.pop();
// 拼接前后节点
if (prev) {
prev->left = nullptr;
prev->right = curr;
}
// 先右后左入栈，弹出顺序为前序
TreeNode* left = curr->left;
TreeNode* right = curr->right;
if (right) stk.push(right);
if (left) stk.push(left);
prev = curr;
}
}
};解法三（前驱结点）
cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void flatten(TreeNode* root) {
TreeNode* curr = root;
while (curr) {
// 存在左子树则重构指针
if (curr->left) {
TreeNode* nxt = curr->left;
TreeNode* predecessor = nxt;
// 遍历到左子树最右端节点
while (predecessor->right) {
predecessor = predecessor->right;
}
// 拼接原右子树
predecessor->right = curr->right;
// 左树迁移到右分支
curr->left = nullptr;
curr->right = nxt;
}
// 向右前进
curr = curr->right;
}
}
};