124 lines
2.8 KiB
Markdown
124 lines
2.8 KiB
Markdown
# Leetcode Validate-Binary-Search-Tree
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#### 2022-07-08 10:36
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> ##### Algorithms:
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>
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> #algorithm #DFS #DFS_inorder
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>
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> ##### Data structures:
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>
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> #DS #binary_tree #binary_search_tree
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>
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> ##### Difficulty:
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>
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> #coding_problems #difficulty_medium
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>
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> ##### Additional tags:
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>
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> #leetcode #CS_list_need_practicing
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/validate-binary-search-tree/)
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***
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### Problem
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Given the `root` of a binary tree, _determine if it is a valid binary search tree (BST)_.
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A **valid BST** is defined as follows:
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- The left subtree of a node contains only nodes with keys **less than** the node's key.
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- ==The right subtree of a node contains only nodes with keys **greater than** the node's key.==
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- Both the left and right subtrees must also be binary search trees.
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#### Examples
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**Example 1:**
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**Input:** root = [2,1,3]
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**Output:** true
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**Example 2:**
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**Input:** root = [5,1,4,null,null,3,6]
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**Output:** false
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**Explanation:** The root node's value is 5 but its right child's value is 4.
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#### Constraints
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- The number of nodes in the tree is in the range `[1, 104]`.
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- `-231 <= Node.val <= 231 - 1`
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### Thoughts
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> [!summary]
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> This is a #DFS #DFS_inorder problem.
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I have thought a lot of recursion methods, but at last I realized:
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> [!tip] The feature of BST
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> For a BST, DFS inorder search returns an array of ascending numbers.
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So, I use a DFS inorder, along with a prev reference pointer to keep track of previous value. to validate, the value of node should always be bigger than prev.
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See comment for why I use a **reference to pointer.**
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### Solution
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
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* right(right) {}
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* };
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*/
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class Solution {
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bool checker(TreeNode *root, int *&prev) {
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// Use reference to pointer here, because the pointer address will change
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// when assigning memory. Also can use pointer to pointer, or initialize the
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// pointer and never change the address ever.
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if (!root) {
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return true;
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}
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if (!checker(root->left, prev)) {
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return false;
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}
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if (prev && root->val <= *prev) {
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return false;
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}
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if (!prev) {
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// prev's address got changed
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prev = new int;
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}
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*prev = root->val;
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return checker(root->right, prev);
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}
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public:
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bool isValidBST(TreeNode *root) {
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// DFS inorder traversal: valid BST returns a ascending array
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int *prev = nullptr;
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return checker(root, prev);
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}
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};
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```
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