# Leetcode Search-In-a-Binary-Tree

#### 2022-07-07 08:06

> ##### Algorithms:
> #algorithm #DFS #BFS
> ##### Data structures:
> #DS #binary_tree 
> ##### Difficulty:
> #coding_problem #difficulty-easy 
> ##### Additional tags:
> #leetcode 
> ##### Revisions:
> N/A

##### Related topics:
```expander
tag:#DFS OR tag:#BFS
```
 
- [[Breadth First Search]]
- [[Leetcode Binary-Tree-Inorder-Traversal]]
- [[Leetcode Binary-Tree-Level-Order-Traversal]]
- [[Leetcode Binary-Tree-Postorder-Traversal]]
- [[Leetcode Binary-Tree-Preorder-Traversal]]
- [[Leetcode Insert-Into-a-Binary-Search-Tree]]
- [[Leetcode Invert-Binary-Tree]]
- [[Leetcode Lowest-Common-Ancestor-Of-a-Binary-Search-Tree]]
- [[Leetcode Maximum-Depth-Of-Binary-Tree]]
- [[Leetcode Path-Sum]]
- [[Leetcode Symmetric-Tree]]
- [[Leetcode Two-Sum-IV-Input-Is-a-BST]]
- [[Leetcode Validate-Binary-Search-Tree]]
 

##### Links:
- [Link to problem](https://leetcode.com/problems/search-in-a-binary-search-tree/submissions/)
___
### Problem

You are given the `root` of a binary search tree (BST) and an integer `val`.

Find the node in the BST that the node's value equals `val` and return the subtree rooted with that node. If such a node does not exist, return `null`.

#### Examples

Example 1:

Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]

Example 2:

Input: root = [4,2,7,1,3], val = 5
Output: []



#### Constraints

-   The number of nodes in the tree is in the range `[1, 5000]`.
-   `1 <= Node.val <= 107`
-   `root` is a binary search tree.
-   `1 <= val <= 107`

### Thoughts

> [!summary]
> This is a #DFS or #BFS problem. We search values in the tree.

In DFS, I use preorder, since the root value will be check first, making it quicker if data appear on shallow trees more.

In BFS, I don't have use for loop inside while loop, since we don't have to consider levels.

### Solution

DFS
```cpp
/**
 * 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 {
public:
  TreeNode *searchBST(TreeNode *root, int val) {
    // DFS preorder
    // Base cases
    if (!root) {
      return nullptr;
    }
    if (root->val == val) {
      return root;
    }

    auto left = searchBST(root->left, val);
    if (left) {
      return left;
    }
    auto right = searchBST(root->right, val);
    if (right) {
      return right;
    }

    // left and right not found
    return nullptr;
  }
};
```

Which can be simplified to 
```cpp
/**
 * 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 {
public:
  TreeNode* searchBST(TreeNode* root, int val) {
    // DFS preorder
    // Base cases
    if (!root) {
      return nullptr;
    }
    if (root->val == val) {
      return root;
    }
    
    auto left = searchBST(root->left, val);
    if (left) {
      return left;
    }
    return searchBST(root->right, val);
  }
};
```

BFS
```cpp
/**
 * 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 {
public:
  TreeNode *searchBST(TreeNode *root, int val) {
    // BFS
    queue<TreeNode *> pending;
    pending.push(root);

    TreeNode *ptr;
    while (!pending.empty()) {
      ptr = pending.front();
      pending.pop();

      if (ptr->val == val) {
        return ptr;
      }

      if (ptr->left)
        pending.push(ptr->left);
      if (ptr->right)
        pending.push(ptr->right);
    }

    return nullptr;
  }
};
```