112. Path Sum

Problem Description

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.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
def hasPathSum(root: Optional[TreeNode], targetSum: int) -> bool:
    if not root:
        return False
    
    # If it's a leaf node, check if current value equals remaining sum
    if not root.left and not root.right:
        return root.val == targetSum
    
    # Recursively check left and right subtrees with reduced target
    return (hasPathSum(root.left, targetSum - root.val) or 
            hasPathSum(root.right, targetSum - root.val))

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) {
            return false;
        }
        
        // If it's a leaf node, check if current value equals remaining sum
        if (root.left == null && root.right == null) {
            return root.val == targetSum;
        }
        
        // Recursively check left and right subtrees with reduced target
        return hasPathSum(root.left, targetSum - root.val) || 
               hasPathSum(root.right, targetSum - root.val);
    }
}

/**
 * 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;
        }
        
        // If it's a leaf node, check if current value equals remaining sum
        if (!root->left && !root->right) {
            return root->val == targetSum;
        }
        
        // Recursively check left and right subtrees with reduced target
        return hasPathSum(root->left, targetSum - root->val) || 
               hasPathSum(root->right, targetSum - root->val);
    }
};

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
var hasPathSum = function(root, targetSum) {
    if (!root) {
        return false;
    }
    
    // If it's a leaf node, check if current value equals remaining sum
    if (!root.left && !root.right) {
        return root.val === targetSum;
    }
    
    // Recursively check left and right subtrees with reduced target
    return hasPathSum(root.left, targetSum - root.val) || 
           hasPathSum(root.right, targetSum - root.val);
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public bool HasPathSum(TreeNode root, int targetSum) {
        if (root == null) {
            return false;
        }
        
        // If it's a leaf node, check if current value equals remaining sum
        if (root.left == null && root.right == null) {
            return root.val == targetSum;
        }
        
        // Recursively check left and right subtrees with reduced target
        return HasPathSum(root.left, targetSum - root.val) || 
               HasPathSum(root.right, targetSum - root.val);
    }
}