112. Path Sum

Solution - Recursion

/**
 * 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:
    bool hasPathSum(TreeNode* root, int targetSum)
    {
        if (!root) return false;
        
        targetSum -= root->val;

        if (!root->left && !root->right) return targetSum == 0;

        return hasPathSum(root->left, targetSum) || hasPathSum(root->right, targetSum);
    }
};

• T:
  $O(N)$
• S:
  $O(N)$

Solution - Iteration

/**
 * 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:
    bool hasPathSum(TreeNode* root, int targetSum)
    {
        if (!root) return false;
        queue<pair<TreeNode*, int>> q;
        q.push({root, targetSum});
        while (!q.empty())
        {
            TreeNode* node = q.front().first;

            int sum = q.front().second; q.pop();
            
            if (!node) continue;
            
            sum -= node->val;
            
            if (sum == 0 && !node->left && !node->right)
            {
                return true;
            }
            
            if (node->left)
            {
                q.push({node->left, sum});
            }
            
            if (node->right)
            {
                q.push({node->right, sum});
            }
        }
        return false;
    }
};

• T:
  $O(N)$
• S:
  $O(N)$