Problem
Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
Note that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.
For example,
1 2 3 4 5 6 7 |
Given the tree: 4 / \ 2 7 / \ 1 3 And the value to insert: 5 |
You can return this binary search tree:
1 2 3 4 5 |
4 / \ 2 7 / \ / 1 3 5 |
This tree is also valid:
1 2 3 4 5 6 7 |
5 / \ 2 7 / \ 1 3 \ 4 |
Solution: Recursion
Time complexity: O(logn ~ n)
Space complexity: O(logn ~ n)
1 2 3 4 5 6 7 8 9 10 11 12 13 |
// Author: Huahua // Running time: 52 ms class Solution { public: TreeNode* insertIntoBST(TreeNode* root, int val) { if (root == nullptr) return new TreeNode(val); if (val > root->val) root->right = insertIntoBST(root->right, val); else root->left = insertIntoBST(root->left, val); return root; } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment