logseq_notes/pages/OJ notes/pages/Leetcode Path-Sum.md

Leetcode Path-Sum

2022-07-06 13:45

Algorithms:

#algorithm #DFS #recursion

Data structures:

#DS #binary_tree

Difficulty:

#coding_problems #difficulty_easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

  ---

Problem

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.

Examples

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 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

Thoughts

[!summary] This is a #DFS recursion problem.

There are one thing to consider, return false when the tree is empty.

Simple DFS-like recursion problem.

Base cases:

• node is empty, return false

• node is leaf

  • if the value is sum, return true

  • else return false

    Pseudo-code:

• Check for base-cases

• return check(left, sum - root->val) || check(right, sum - root->val)

  [!tip] Why use OR By using OR operator, return true when there is at least one solution that matches.

Solution

/**
 * 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) {
  // DFS In-order Recursion

  // Base case: node does not exist
  if (!root) {
    return false;
  }
  int val = root->val;

  // Base case: reached leaf
  if (!root->left && !root->right) {
    if (targetSum == val)
      return true;
    else
      return false;
  }

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