112. Path Sum

Recursion

/*
 * @lc app=leetcode id=112 lang=cpp
 *
 * [112] Path Sum
 *
 * https://leetcode.com/problems/path-sum/description/
 *
 * algorithms
 * Easy (51.70%)
 * Likes:    10144
 * Dislikes: 1170
 * Total Accepted:    1.8M
 * Total Submissions: 3.4M
 * Testcase Example:  '[5,4,8,11,null,13,4,7,2,null,null,null,1]\n22'
 *
 * Given the root of a binary tree and an integer targetSum, return true if the
 * tree has a root-to-leaf path such that adding up all the values along the
 * path equals targetSum.
 *
 * A leaf is a node with no children.
 *
 *
 * Example 1:
 *
 *
 * Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
 * Output: true
 * Explanation: The root-to-leaf path with the target sum is shown.
 *
 *
 * Example 2:
 *
 *
 * Input: root = [1,2,3], targetSum = 5
 * Output: false
 * Explanation: There are two root-to-leaf paths in the tree:
 * (1 --> 2): The sum is 3.
 * (1 --> 3): The sum is 4.
 * There is no root-to-leaf path with sum = 5.
 *
 *
 * Example 3:
 *
 *
 * Input: root = [], targetSum = 0
 * Output: false
 * Explanation: Since the tree is empty, there are no root-to-leaf paths.
 *
 *
 *
 * Constraints:
 *
 *
 * The number of nodes in the tree is in the range [0, 5000].
 * -1000 <= Node.val <= 1000
 * -1000 <= targetSum <= 1000
 *
 *
 */

// @lc code=start
/**
 * 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);
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$

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: $O(N)$