133 lines
2.8 KiB
Markdown
133 lines
2.8 KiB
Markdown
# Leetcode Path-Sum
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#### 2022-07-06 13:45
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> ##### Algorithms:
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>
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> #algorithm #DFS #recursion
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>
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> ##### Data structures:
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>
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> #DS #binary_tree
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty-easy
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/path-sum/)
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---
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### Problem
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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`.
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A **leaf** is a node with no children.
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#### Examples
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg)
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**Input:** root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
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**Output:** true
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**Explanation:** The root-to-leaf path with the target sum is shown.
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg)
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**Input:** root = [1,2,3], targetSum = 5
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**Output:** false
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**Explanation:** There two root-to-leaf paths in the tree:
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(1 --> 2): The sum is 3.
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(1 --> 3): The sum is 4.
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There is no root-to-leaf path with sum = 5.
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**Example 3:**
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**Input:** root = [], targetSum = 0
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**Output:** false
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**Explanation:** Since the tree is empty, there are no root-to-leaf paths.
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#### Constraints
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- The number of nodes in the tree is in the range `[0, 5000]`.
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- `-1000 <= Node.val <= 1000`
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- `-1000 <= targetSum <= 1000`
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### Thoughts
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> [!summary]
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> This is a #DFS recursion problem.
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There are one thing to consider, return false when the tree is empty.
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Simple DFS-like recursion problem.
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Base cases:
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- node is empty, return false
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- node is leaf
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- if the value is sum, return true
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- else return false
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Pseudo-code:
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- Check for base-cases
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- return check(left, sum - root->val) || check(right, sum - root->val)
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> [!tip] Why use OR
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> By using OR operator, return true when there is at least one solution that matches.
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### Solution
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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bool hasPathSum(TreeNode* root, int targetSum) {
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// DFS In-order Recursion
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// Base case: node does not exist
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if (!root) {
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return false;
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}
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int val = root->val;
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// Base case: reached leaf
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if (!root->left && !root->right) {
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if (targetSum == val)
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return true;
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else
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return false;
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}
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return hasPathSum(root->left, targetSum - val) || hasPathSum(root->right, targetSum - val);
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}
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};
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```
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