logseq_notes/pages/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_problems #difficulty_easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

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;
}
};