Problem

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Problem Solving Approach

• The objective of this problem is to check whether there is a path in the given binary tree from the root to a leaf node whose sum equals the given targetSum.

• Therefore, this code recursively explores paths in the given binary tree and checks if the path sum matches the targetSum.

Algorithm

1. Recursion:

   • The hasPathSum function is called recursively.

   • This function subtracts the value of the current node from targetSum and uses the result as the new targetSum.

2. Base Case:

   • At the start of the recursive call, it checks if the current node is null to determine if it has reached the end of the tree.

   • If it's null, the current path is not valid, so it returns false.

3. Leaf Node Check:

   • It checks if the current node is a leaf node, meaning it has no children.

   • If the current node is a leaf node, it checks if targetSum is equal to 0.

   • If targetSum is 0, it means that the current path's sum matches the targetSum, so it returns true.

4. Recursive Calls for Left and Right Subtrees:

   • If the current node is not a leaf node, it makes recursive calls to the left and right subtrees using the modified targetSum value.

5. Return Result:

   • If either the left or right subtree returns true, it means there exists a path in the current path that satisfies the targetSum, so it returns true.

   • Otherwise, it returns false.

Recursive Function Implementation Table

• Goal: https://leetcode.com/problems/path-sum/

• Base Case (Termination Conditions):

    if root == null
    return false
  
    if left == null && right == null <- after subtracting root.val from sum
    return sum == 0
  

• Is the previous step's result needed?: Yes

• Problem Splitting (Divide the Problem):

    sum -= root.val
  

• Combining Results:

    boolean left
    boolean right
    return left || right
  

• Recursive Calls and Changes Before Moving to the Next Step (Recursive Call):

    sum -= root.val
  

Github Link

https://github.com/eunhanlee/LeetCode_112_PathSum_Solution.git

Time Complexity: O(n), Space Complexity: O(h)

class Solution {

    /**
     * Checks if there is a path from the root to a leaf node in the given binary tree
     * with a sum equal to the target sum.
     *
     * @param root      The root node of the binary tree.
     * @param targetSum The target sum to be achieved.
     * @return Returns `true` if a path exists, `false` otherwise.
     */
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) {
            return false;
        }

        // Subtract the value of the current node from the target sum.
        targetSum -= root.val;

        // Check if the current node is a leaf node and if the target sum is 0.
        if (root.left == null && root.right == null) {
            return targetSum == 0;
        }

        // Recursively check the left and right subtrees.
        Boolean rightNode = hasPathSum(root.right, targetSum);
        Boolean leftNode = hasPathSum(root.left, targetSum);

        // Return `true` if there is a valid path in either the left or right subtree.
        return rightNode || leftNode;
    }
}

Time Complexity

• The time complexity of this code is O(n) because it visits all nodes in the tree exactly once, where n is the number of nodes in the tree.

Space Complexity

• Due to the recursive function calls, a call stack is built up, but it only requires space up to the height of the tree, so the space complexity is O(h), where h is the height of the tree.