Description:
Given a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.
Example:
Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]
5
/ \
3 6
/ \ \
2 4 8
/ / \
1 7 9
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
1
\
2
\
3
\
4
\
5
\
6
\
7
\
8
\
9
Note:
The number of nodes in the given tree will be between 1 and 100.
Each node will have a unique integer value from 0 to 1000.
题目描述:
给定一个树，按中序遍历重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。
示例 :
输入：[5,3,6,2,4,null,8,1,null,null,null,7,9]
5
/ \
3 6
/ \ \
2 4 8
/ / \
1 7 9
输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
1
\
2
\
3
\
4
\
5
\
6
\
7
\
8
\
9
提示：
给定树中的结点数介于 1 和 100 之间。
每个结点都有一个从 0 到 1000 范围内的唯一整数值。
思路:
- 使用栈按中序遍历存储结点
- 递归法
- 步骤: 记录子树的父结点, 如果该结点是左结点, 记录的是其父结点, 否则记录的是父结点的父结点
- 原理: 按照中序遍历, 对于现在遍历的结点: 如果是左结点, 那么对于结果链表(result linked list)来说其下一个结点应该是父结点; 如果是右结点, 那么对于结果链表来说其上一个结点是父结点, 下一个结点应该是父结点的父结点(即要遍历完父结点为根的子树)
- in_order(root.left) + root + in_order(right), root -> left = nullptr, root -> right = next_node(注意在这一步, 该结点的父结点已经遍历完成)
- 总的来说, 最后返回的, 如果不考虑左结点(nullptr), 可以看作一个链表, 其关键字是按照升序排列的
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* increasingBST(TreeNode* root)
{
stack s;
TreeNode* result = new TreeNode(0);
TreeNode* p = result;
while (s.size() || root)
{
if (root)
{
s.push(root);
root = root -> left;
}
else
{
TreeNode* cur = s.top();
s.pop();
root = cur -> right;
cur -> left = nullptr;
p -> right = cur;
p = p -> right;
}
}
return result -> right;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode increasingBST(TreeNode root) {
return helper(root, null);
}
private TreeNode helper(TreeNode root, TreeNode temp) {
if (root == null) return temp;
TreeNode result = helper(root.left, root);
root.left = null;
root.right = helper(root.right, temp);
return result;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def increasingBST(self, root: TreeNode) -> TreeNode:
def helper(root: TreeNode, temp: TreeNode=None) -> TreeNode:
if not root:
return temp
result, root.left, root.right = helper(root.left, root), None, helper(root.right, temp)
return result
return helper(root)