814. Binary Tree Pruning

Version #2 DFS -> using null to represent false

47.32 %
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        if (root == null) return null;
        root.left = pruneTree(root.left);
        root.right = pruneTree(root.right);
        if (root.left == null && root.right == null && root.val == 0) return null;
        return root;
    }
}


Version #1 Typical DFS

5.28 %
class Solution {
    public TreeNode pruneTree(TreeNode root) {
        // recursively check if a node's left&right subtree containing 1
        // if substree returns false, then set that subtree to null
        // return dfs(left) || dfs(right) || node.val == 1
        if (dfs(root)) return root;
        return null;
    }
    private boolean dfs(TreeNode node) {
        if (node == null) return false;
        boolean left = dfs(node.left);
        boolean right = dfs(node.right);
        if (!left) {
            node.left = null;
        }
        if (!right) {
            node.right = null;
        }
        return left || right || node.val == 1;
    }
}