Java DFS 8ms Easy Solution


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    Simple preorder traversal with a boolean parameter l indicating whether a left child or not

    class Solution {
        public int sumOfLeftLeaves(TreeNode root) {
            return preorder(root, false);
        }
        
        int preorder(TreeNode root, boolean l){
            if(root == null) return 0;
            int sum = 0;
            if(root.left == null && root.right == null && l) sum+=root.val;
            return sum+preorder(root.left, true)+preorder(root.right, false);
        }
    }
    

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