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Path Sum

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

112. Path Sum

Problem Description

Given the root of a binary tree and an integer targetSum, determine if there exists 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.

Solution: Depth-First Search (DFS)

The most efficient approach to solving this problem is using Depth-First Search (DFS). DFS recursively explores the tree, checking each path from the root to a leaf.

Algorithm:

  1. Base Case: If the current node is null, there's no path, so return false.

  2. Leaf Node Check: If the current node is a leaf node (both left and right children are null), check if the current path sum (currentSum) equals the targetSum. If they are equal, return true; otherwise, return false.

  3. Recursive Calls: Recursively call the DFS function on the left and right subtrees, updating the currentSum by adding the current node's value. If either recursive call returns true, it means a path with the target sum was found, so return true. Otherwise, return false.

Time Complexity: O(N), where N is the number of nodes in the tree. We visit each node at most once.

Space Complexity: O(H), where H is the height of the tree. This is due to the recursive call stack. In the worst case (a skewed tree), H can be equal to N.

Code Implementation

The following code implements the DFS solution in several programming languages:

Python:

# 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
 
class Solution:
    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        def dfs(node, currentSum):
            if not node:
                return False
            currentSum += node.val
            if not node.left and not node.right:  # Leaf node
                return currentSum == targetSum
            return dfs(node.left, currentSum) or dfs(node.right, currentSum)
 
        return dfs(root, 0)

Java:

/**
 * 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) {
        return dfs(root, targetSum, 0);
    }
 
    private boolean dfs(TreeNode node, int targetSum, int currentSum) {
        if (node == null) {
            return false;
        }
        currentSum += node.val;
        if (node.left == null && node.right == null) { // Leaf node
            return currentSum == targetSum;
        }
        return dfs(node.left, targetSum, currentSum) || dfs(node.right, targetSum, currentSum);
    }
}

C++:

/**
 * 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) {
        return dfs(root, targetSum, 0);
    }
 
    bool dfs(TreeNode* node, int targetSum, int currentSum) {
        if (!node) return false;
        currentSum += node->val;
        if (!node->left && !node->right) return currentSum == targetSum;
        return dfs(node->left, targetSum, currentSum) || dfs(node->right, targetSum, currentSum);
    }
};

Go:

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func hasPathSum(root *TreeNode, targetSum int) bool {
	return dfs(root, targetSum, 0)
}
 
func dfs(node *TreeNode, targetSum, currentSum int) bool {
	if node == nil {
		return false
	}
	currentSum += node.Val
	if node.Left == nil && node.Right == nil { //Leaf node
		return currentSum == targetSum
	}
	return dfs(node.Left, targetSum, currentSum) || dfs(node.Right, targetSum, currentSum)
}

//Similar implementations can be provided for other languages like JavaScript, TypeScript, and Rust, following the same logical structure. The core idea remains consistent across all languages.