96 lines
2.4 KiB
Markdown
96 lines
2.4 KiB
Markdown
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# Leetcode Insert-Into-a-Binary-Search-Tree
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#### 2022-07-07 08:50
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> ##### Algorithms:
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> #algorithm #recursion #DFS
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> ##### Data structures:
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> #DS #binary_tree
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> ##### Difficulty:
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> #coding_problem #difficulty-medium
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#DFS
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```
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- [[Leetcode Binary-Tree-Inorder-Traversal]]
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- [[Leetcode Binary-Tree-Postorder-Traversal]]
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- [[Leetcode Binary-Tree-Preorder-Traversal]]
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- [[Leetcode Invert-Binary-Tree]]
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- [[Leetcode Maximum-Depth-Of-Binary-Tree]]
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- [[Leetcode Path-Sum]]
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- [[Leetcode Search-In-a-Binary-Tree]]
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- [[Leetcode Symmetric-Tree]]
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##### Links:
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- [Link to problem](https://leetcode.com/problems/insert-into-a-binary-search-tree/)
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___
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### Problem
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You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
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Notice 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.
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#### Examples
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**Example 1:**
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**Input:** root = [4,2,7,1,3], val = 5
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**Output:** [4,2,7,1,3,5]
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**Explanation:** Another accepted tree is:
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**Example 2:**
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**Input:** root = [40,20,60,10,30,50,70], val = 25
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**Output:** [40,20,60,10,30,50,70,null,null,25]
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**Example 3:**
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**Input:** root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
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**Output:** [4,2,7,1,3,5]
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#### Constraints
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- The number of nodes in the tree will be in the range `[0, 104]`.
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- `-108 <= Node.val <= 108`
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- All the values `Node.val` are **unique**.
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- `-108 <= val <= 108`
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- It's **guaranteed** that `val` does not exist in the original BST.
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### Thoughts
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> [!summary]
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> This is a #DFS problem.
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Seems that no one seems to care about the second way, since it's way too complex.
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#### DFS
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DFS-like solution is a simple recursion problem, using a helper function.
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##### Edge case:
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root is null, simple return a new node.
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##### Base cases:
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- left side is empty, and val < root->val: place it in
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- right side is empty, and val > root->val: place it in
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##### Pseudocode:
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- check for base cases
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- if val < root->val, insert(root->left, val)
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- vice, versa.
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#### BFS
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### Solution
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