在二叉树中查找节点的父节点

Question

I am trying to write a method to find the parent of a given node. Here's my method.
I created a BinaryNode object r which initially refers to root.

    public BinaryNode r=root;
    public BinaryNode parent(BinaryNode p){
    BinaryNode findParent=p;        
        if (isRoot(findParent) || r==null){
                return null;
        }
        else{
            if(r.left==findParent || r.right==findParent)
                return r;
            else{
                if (r.element<findParent.element)
                    return parent(r.right);
                else
                    return parent(r.left);
            }
        }
    }  

THis code doesn't work properly .I think that's because r is a null object.Because when I do

if (isRoot(findParent) || r==null){
                System.out.println(r==null);
                return null;}  

r==null evaluates to true .How come that happen because I have inserted nodes as

public static void main (String args[]){
        BinaryTree t=new BinaryTree();
        t.insert(5);
        t.insert(t.root,4);
        t.insert(t.root,6);
        t.insert(t.root,60);
        t.insert(t.root,25);
        t.insert(t.root,10);  

and the root is not null.
Can some one please point out why that happens and if what I am trying to do in order to find the parent node is logically correct.

Answer 1

The problem is that you MUST keep track of your current node, while keeping the node who's parent you want to find. And as far as I understand your code, you keep the variable, but never change it
I'd recommend using a helper function. This would look something like that:

public BinaryNode parent(BinaryNode p){
    parentHelper(root,p)
}
private BinaryNode parentHelper(BinaryNode currentRoot, BinaryNode p) {        
    if (isRoot(p) || currentRoot==null){
            return null;
    }
    else{
        if(currentRoot.left==p || currentRoot.right==p)
            return currentRoot;
        else {
            if (currentRoot.element<p.element) {
                return parentHelper(currentRoot.right,p);
            }
            else {
                return parentHelper(currentRoot.left,p);
            }
        }
    }
}  

Answer 2

I compared value to value because I didn't define the way to compare the nodes.

    public static Node FindParent(Node root, Node node)
    {
        if (root == null || node == null)
        {
            return null;
        }
        else if ( (root.Right != null && root.Right.Value == node.Value) || (root.Left != null && root.Left.Value == node.Value))
        {
            return root;
        }
        else
        {
            Node found = FindParent(root.Right, node);
            if (found == null)
            {
                found = FindParent(root.Left, node);
            }
            return found;
        }
    }

Answer 3

使用两个参数：一个用于当前节点，一个用于正在搜索的节点。

Answer 4

Here is code to find out parent node using a stack Data Structures.

Stack<TreeNode> parentStack = new Stack<TreeNode>();

public static void inOrderTraversal(TreeNode root){
            if(root != null){
                if(parentStack.size()==0){
                    parentStack.push(root);
                }
                if(root.getLeftChild()!=null){
                    parentStack.push(root); 
                    inOrderTraversal(root.getLeftChild());  
                }
                parent = parentStack.pop();
                System.out.println(root.getNodeValue()+"'s parent is "+parent.getNodeValue());
                if(root.getRightChild()!=null){
                    parentStack.push(root);
                    inOrderTraversal(root.getRightChild());
                }
            }
            else{
                if(root==null){System.err.println("Can't process a empty root tree");}
            }
        }

Answer 5

I prefer to delegate as much work as I can to lower-level components, in this case the Node class. Re: finding a node's parent, this is how I do it ...

template <typename T>
Node<T>* Node<T>::parent (const Node<T>* node) 
{
    if (node)
    {
        if (*node < *this)
        {
            if (left && (*node < *left))
                return left->parent (node);
            return this;
        }
        if (*node > *this)
        {
            if (right && (*right > *node))
                return right->parent (node);
            return this;
        }
    }
    return nullptr;
}