Problem
Given a 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 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.
Solution: In-order traversal
root = 5
inorder(root.left) 之后
self.prev = 4
(1-4)已经处理完了,这时候的树是很奇怪的一个形状, 3即是2的右子树,又是5的左子树。
1
\
2 5
\ / \
3 6
\ \
prev -> 4 8
/ \
7 9
—————————
5.left = None # 把5->3的链接断开
5
\
6
\
8
/ \
7 9
—————————–
self.prev.right = root <=> 4.right = 5
把5接到4的右子树
1
\
2
\
3
\
4
\
5 <– prev
\
6
\
8
/ \
7 9
self.prev = root <=> prev = 5
inorder(5.right) <=> inorder(6) 然后再去递归处理6(及其子树)即可。
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua // Running time: 64 ms class Solution { public: TreeNode* increasingBST(TreeNode* root) { TreeNode dummy(0); prev_ = &dummy; inorder(root); return dummy.right; } private: TreeNode* prev_; void inorder(TreeNode* root) { if (root == nullptr) return; inorder(root->left); prev_->right = root; prev_ = root; prev_->left = nullptr; inorder(root->right); } }; |
Java
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class Solution { private TreeNode prev; public TreeNode increasingBST(TreeNode root) { TreeNode dummy = new TreeNode(0); this.prev = dummy; inorder(root); return dummy.right; } private void inorder(TreeNode root) { if (root == null) return; inorder(root.left); this.prev.right = root; this.prev = root; this.prev.left = null; inorder(root.right); } } |
Python3
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class Solution: def increasingBST(self, root): dummy = TreeNode(0) self.prev = dummy def inorder(root): if not root: return None inorder(root.left) root.left = None self.prev.right = root self.prev = root inorder(root.right) inorder(root) return dummy.right |
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