404. Sum of Left Leaves (Easy)

Find the sum of all left leaves in a given binary tree.

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

Solution 1: DFS O(n); O(h)

    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int sum = 0; //sum should be declared as a global variable.
        //check if current node is left leaf.
        if (root.left != null && root.left.left == null && root.left.right == null) {
            sum += root.left.val;
        }
        sum += sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
        return sum;
    }

Solution 2: BFS O(n); O(w)

    public int sumOfLeftLeaves(TreeNode root) {
        if(root == null || root.left == null && root.right == null) return 0;

        int res = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while(!queue.isEmpty()) {
            TreeNode curr = queue.poll();

            if(curr.left != null && curr.left.left == null && curr.left.right == null) res += curr.left.val;
            if(curr.left != null) queue.offer(curr.left);
            if(curr.right != null) queue.offer(curr.right);
        }
        return res;
    }