Leetcode112.Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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Approach1: DFS

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This problem is quite similar to the LeetCode104, I use two stacks, one for node and one for node value. This is a pre-order idea to process nodes. Each time when I process the node, I process the value of it and update the sum of it. Cause the problem is asked about the root-leaf. I thought a valid path is right. that's the mistake I made at first. So I need to check if the node has child even if the value is equal to sum.

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        Stack stack = new Stack<>();
        Stack value = new Stack<>();
        stack.push(root);
        value.push(root.val);
        while(!stack.empty()){
            TreeNode node = stack.pop();
            int temp = value.pop();
            if(temp == sum && node.left == null && node.right == null) {//follow up may exist here
                return true;
            }
            if(node.left != null){
                stack.push(node.left);
                value.push(node.left.val + temp);
            }
            if(node.right != null){
                stack.push(node.right);
                value.push(node.right.val + temp);
            }     
        }
        return false;
    }
}

Follow up question my be like: Find a valid path is fine, no need to root-leaf. Then just remove the condition "node.left == null && node.right == null".

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Approach2 Recursion

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The basic idea is to subtract the value of current node from sum until it reaches a leaf node and the subtraction equals 0, then we know that we got a hit. Otherwise the subtraction at the end could not be 0. Recursion pattern goes like update the parameter while call it self.

 class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
    
        if(root.left == null && root.right == null && sum - root.val == 0) return true;
    
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}