二叉树中节点到根路径

Question

I was trying to find the node to root path in a binary tree using Leap Of Faith but I'm stuck now.

It works in some simple cases, but it returns the wrong path for other cases.

For example, for this tree:

        1
       / \
      2   3

...and passing 3 as argument, it returns {1, 2} when {1, 3} is expected.

Can anyone tell me what is wrong with my code?

public static ArrayList<Integer> nodeToRootPath(Node root, int data) {
    // write your code here
    ArrayList<Integer> res = new ArrayList<>();
    if (root == null) {
        return res;
    }
    if (root.data == data) {
        res.add(data);
        return res;
    }
    res = nodeToRootPath(root.left, data);
    if (res.size() != 0) {
        res.add(root.data);
        return res;
    }
    else {
        res.clear();
        res = nodeToRootPath(root.right, data);
        res.add(root.data);
        return res;
    }
}

Answer 1

The problem is that when the recursion comes back from the right subtree, your code does not verify whether that search was successful. It should -- just like for the left-side -- check whether res.size() is non-zero before adding anything to it.

Some side notes:

• It is not necessary to call res.clear() when you are going to assign a new value to res in the next line.
• It is pity to create a new list when eventually you are not in a base case and ignore that created new list and instead use the list you get back from a recursive call. So only create a new list when you know you are in a base case.

Here is a correction:

public static ArrayList<Integer> nodeToRootPath(Node root, int data) {
    if (root == null) {
        return new ArrayList<>();
    }
    if (root.data == data) {
        return new ArrayList<>(Arrays.asList(data));
    }
    ArrayList<Integer> res = nodeToRootPath(root.left, data);
    if (res.size() == 0) {
        res = nodeToRootPath(root.right, data);
    }
    if (res.size() != 0) {
        res.add(root.data);
    }
    return res;
}