logseq_notes/pages/OJ notes/pages/Leetcode Validate-Binary-Search-Tree.md

Leetcode Validate-Binary-Search-Tree

2022-07-08 10:36

Algorithms:

#algorithm #DFS #DFS_inorder

Data structures:

#DS #binary_tree #binary_search_tree

Difficulty:

#coding_problems #difficulty_medium

Additional tags:

#leetcode #CS_list_need_practicing

Revisions:

N/A

Related topics:

Links:

• Link to problem

  ---

Problem

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• ==The right subtree of a node contains only nodes with keys greater than the node's key.==
• Both the left and right subtrees must also be binary search trees.

Examples

Example 1:

Input: root = [2,1,3] Output: true

Example 2:

Input: root = [5,1,4,null,null,3,6] Output: false Explanation: The root node's value is 5 but its right child's value is 4.

Constraints

• The number of nodes in the tree is in the range [1, 104].
• -231 <= Node.val <= 231 - 1

Thoughts

[!summary] This is a #DFS #DFS_inorder problem.

I have thought a lot of recursion methods, but at last I realized:

[!tip] The feature of BST For a BST, DFS inorder search returns an array of ascending numbers.

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.

See comment for why I use a reference to pointer.

Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
bool checker(TreeNode *root, int *&prev) {
  // Use reference to pointer here, because the pointer address will change
  // when assigning memory. Also can use pointer to pointer, or initialize the
  // pointer and never change the address ever.
  if (!root) {
    return true;
  }
  if (!checker(root->left, prev)) {
    return false;
  }
  if (prev && root->val <= *prev) {
    return false;
  }
  if (!prev) {
    // prev's address got changed
    prev = new int;
  }
  *prev = root->val;
  return checker(root->right, prev);
}

public:
bool isValidBST(TreeNode *root) {
  // DFS inorder traversal: valid BST returns a ascending array
  int *prev = nullptr;
  return checker(root, prev);
}
};