Find the diameter of a binary tree

Question

I am trying to find the diameter of a binary tree (Path length between any two nodes in the tree containing maximum number of nodes.) in java.

my code snippet:

Answer 1

You should use the height of the tree to calculate the diameter. Make a getHeight() function which will take the root of the tree as argument and return the height of the tree. Using this value and with the help of recursion we can calculate the diameter of the tree. Here is the code for it.....

Function for calculating the diameter :-

public static int getDiameter(BinaryTreeNode root) {        
  if (root == null)
    return 0;

int rootDiameter = findHeight(root.getLeft()) + findHeight(root.getRight()) + 1;
int leftDiameter = getDiameter(root.getLeft());
int rightDiameter = getDiameter(root.getRight());

return Math.max(rootDiameter, Math.max(leftDiameter, rightDiameter));
}

Function for calculating the height of tree:-

public static int findHeight(BinaryTreeNode node) {
if(node == null)
    return 0;

else {
    return 1+Math.max(findHeight(node.left), findHeight(node.right));
}
}

Answer 2

Algo takes O(n). Calculates height and path at the same time.

public static int findLongestPath(TreeNode root)
{
  // longest path = max (h1 + h2 + 2, longestpath(left), longestpath(right);

  int[] treeInfo = longestPathHelper(root);

  return treeInfo[0];
}

private static int[] longestPathHelper(TreeNode root)
{
  int[] retVal = new int[2];

  if (root == null)
  {
     //height and longest path are 0
     retVal[0] = 0;
     retVal[1] = 0;
  }

  int[] leftInfo = longestPathHelper(root.getLeft());
  int[] rightInfo = longestPathHelper(root.getRight());

  retVal[0] = Math.max(leftInfo[1] + rightInfo[1] + 2, Math.max(leftInfo[0], rightInfo[0]));
  retVal[1] = Math.max(leftInfo[1], rightInfo[1]) + 1;

  return retVal;
}

Answer 3

public int diameter (Node root)
{
    if (root == null) return 0;
    else return Math.max (
        diameter (root.left), 
        Math.max (
            diameter (root.right),
            height (root.left) + height (root.right) + 1));
}

public int height (Node root)
{
    if (root == null) return 0;
    else return 1 + Math.max (height (root.left), height (root.right));
}

Answer 4

I suggest the following:

public static TreeAttr calcTreeDiameter(Node root) {
    if (root == null)
        return new TreeAttr(0, 0);

    TreeAttr leftAttr = calcTreeDiameter(root.getLeft());
    TreeAttr rightAttr = calcTreeDiameter(root.getRight());

    int maxDepth = Math.max(leftAttr.depth, rightAttr.depth);
    int maxDiam = Math.max(leftAttr.diameter, rightAttr.diameter);
    maxDiam = Math.max(maxDiam, leftAttr.depth + rightAttr.depth + 1);

    return new TreeAttr(maxDiam, maxDepth + 1);
}

The TreeAttr is a simple structure containing the diameter and depth of a subtree. Both should be passed in the recursion, since the optimum may either come from one of the subtrees, or from the concatenation of the longest paths.

Answer 5

int max=0;
public int diameter(Tree root) {
  if(root==null) return 0;
  int l=diameter(root.left);
  int r=diameter(root.right);
  max=Math.max(max,l+r+1);
  return l>r:l+1:r+1;
}

max is the max diameter.

Answer 6

You find a diameter of a tree by running a BFS from any node and then another BFS from the furthest node(the node last visited during the first BFS). The diameter is formed by the node last visited in the first BFS and the node last visited in the first BFS. The fact that the tree is binary does not affect the algorithm.

EDIT: changing the value of d in the code you have written will not affect the argument you pass as primitive types are not passed by reference in java.