114 Flatten Binary Tree to Linked List 二叉树展开为链表
Description:
Given a binary tree, flatten it to a linked list in-place.
Example:
For example, given the following tree:
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
题目描述:
给定一个二叉树，原地将它展开为一个单链表。
示例 :
例如，给定二叉树
1
/ \
2 5
/ \ \
3 4 6
将其展开为：
1
\
2
\
3
\
4
\
5
\
6
思路:
后序遍历
将右指针指向左子树
左子树的最后节点的右指针指向原来的右子树
将左指针置空
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
/**
* 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;
flatten(root -> left);
flatten(root -> right);
TreeNode* temp = root -> right;
root -> right = root -> left;
root -> left = nullptr;
while (root -> right) root = root -> right;
root -> right = temp;
}
};
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;
flatten(root.left);
flatten(root.right);
TreeNode temp = root.right;
root.right = root.left;
root.left = null;
while (root.right != null) root = root.right;
root.right = temp;
}
}
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: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
s = []
while root or s:
while root:
s.append(root)
root = root.left
if s:
cur = s.pop()
temp, cur.right, cur.left = cur.right, cur.left, None
while cur.right:
cur = cur.right
cur.right, root = temp, temp
