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:

 graph TD;
     A[5] --> B[4] & C[8]
     B --> D[11] & E[null]
     C --> F[13] & G[4]
     D --> H[7] & I[2]
     G --> L[5] & M[1]

style A fill:#f9f
style B fill:#f9f
style D fill:#f9f
style I fill:#f9f
  
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Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: Following root-to-leaf path exists: [5,4,11,2], marked with () tree. There can be multiple paths, but we have to just return boolean value for that.

Example 2:

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  1
 /  \
2    3

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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:

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Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

Solution

Method 1 - Using Recursion

Code

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public boolean hasPathSum(TreeNode root, int targetSum) {
	if(root == null){
		return false;
	}
	if(root.left == null && root.right == null){
		return root.val == targetSum;
	}
	targetSum = targetSum - root.val;
	return hasPathSum(root.left, targetSum) || hasPathSum(root.right, targetSum)
}

Method 2 - Using Queue

Code

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public boolean hasPathSum(TreeNode root, int sum) {
	if(root == null) return false;

	Queue<TreeNode> nodes = new LinkedList<TreeNode>();
	Queue<Integer> values = new LinkedList<Integer>();

	nodes.add(root);
	values.add(root.val);

	while(!nodes.isEmpty()){
		TreeNode curr = nodes.poll();
		int sumValue = values.poll();

		if(curr.left == null && curr.right == null && sumValue==sum){
			return true;
		}

		if(curr.left != null){
			nodes.add(curr.left);
			values.add(sumValue + curr.left.val);
		}

		if(curr.right != null){
			nodes.add(curr.right);
			values.add(sumValue + curr.right.val);
		}
	}

	return false;
}