notes/OJ notes/pages/Leetcode Search-In-a-Binary-Tree.md

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:

tag:#DFS OR tag:#BFS

• 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 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

---

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

/**
 * 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

/**
 * 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

/**
 * 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;
  }
};