Return parent of node in Binary Tree

问题

I'm writing a code to return the parent of any node, but I'm getting stuck. I don't want to use any predefined ADTs.

//Assume that nodes are represented by numbers from 1...n where 1=root and even 
//nos.=left child and odd nos=right child.
public int parent(Node node){
    if (node % 2 == 0){
       if (root.left==node)
       return root;
    else
       return parent(root.left);
    }
    //same case for right
}

But this program is not working and giving wrong results. My basic algorithm is that the program starts from the root checks if it is on left or on the right. If it's the child or if the node that was queried else, recurses it with the child.

回答1:

This could be rephrased as traverse a binary tree to find a node that is parent to the given one.

Suppose you have a

class Node {
  int node;
  Node left;
  Node right;

  Node(int node, Node left, Node right) {
    this.node = node;
    this.left = left;
    this.right = right;
  }
  @Override
  public String toString (){
     return "("+node+")";
  }
}

For simplicity we will just define global variables.

Node root;
int target;
boolean found;

They will be accessed by the next methods. First, we initialize a method call

public Node findParent(int target){
  found = false;
  this.target = target;
  return internalFindParent(root, null);
}

Second, we write an implementation

private Node internalFindParent(Node node, Node parent){
  if (found) return parent;
  if (node.node == target) {
    found = true;
    return parent;
  }
  if (node.left == null) return null;
  Node temp = internalFindParent(node.left, node);
  if(temp != null)
    return temp;
  if (node.right == null) return null;
  temp = internalFindParent(node.right, node);
  if(temp != null)
    return temp;
  return null;
}

This method traverses a tree and returns results immediately if the given node is found. To demonstrate how it's worked we should create a sample tree and assign it to root node. We numerate each node with the unique number used as a target.

public void init() {
  root = new Node (0,
    new Node(1,
      new Node (2,
        new Node (3,
          new Node (4, null, null),
          new Node (5, null, null)
        ),
        new Node (6,
          new Node (7, null, null),
          new Node (8, null, null)
        )
      ),
      new Node (9,
        new Node (10,
          new Node (11, null, null),
          new Node (12, null, null)
        ),
        new Node (13,
          new Node (14, null, null),
          new Node (15, null, null)
        )
      )
     ),
    new Node(21,
      new Node (22,
        new Node (23,
          new Node (24, null, null),
          new Node (25, null, null)
        ),
        new Node (26,
          new Node (27, null, null),
          new Node (28, null, null)
        )
      ),
      new Node (29,
        new Node (30,
          new Node (31, null, null),
          new Node (32, null, null)
        ),
        new Node (33,
          new Node (34, null, null),
          new Node (35, null, null)
        )
      )
    )
  );
}

Just do all tests in the constructor for simplicity

FindingParent(){
  init();
  for (int i=0; i<=35; i++){
    Node parent = findParent(i);
    if (parent != null)
      System.out.println("("+parent.node+", "+i+")");
  }

}
/**
 * @param args
 */
public static void main(String[] args) {
  new FindingParent();
  System.exit(0);
}

This output results as pairs of (parent, child) for each node in the tree.

回答2:

Try this .It may work :

public BinaryTreeNode getParent(BinaryTreeNode root, BinaryTreeNode node) {

    BinaryTreeNode lh = null, rh = null;
    if (null == root)
        return null;

    if (root.getLeft() == node || root.getRight() == node)
        return root;

    lh = getParent(root.getLeft(), node);
    rh = getParent(root.getRight(), node);

    return lh != null ? lh : rh;

}

来源：https://stackoverflow.com/questions/12342131/return-parent-of-node-in-binary-tree