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Code Binary Tree

Count complete odd nodes paths of binary tree in java

All odd nodes path from root to leaf in binary tree

Java program for Count complete odd nodes paths of binary tree. Here mentioned other language solution.

/* 
  Java program for
  Count odd paths in Binary Tree
*/
// Binary Tree node
class TreeNode
{
	public int data;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(int data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	public TreeNode root;
	public BinaryTree()
	{
		//Set initial tree root 
		this.root = null;
	}
	// Count all paths from root to leaf 
	// which containing all Odd nodes
	public int countOddNodePath(TreeNode node)
	{
		if (node == null || node.data % 2 == 0)
		{
			// When tree node is null or 
			// its contain even value
			return 0;
		}
		else
		{
			if (node.left == null && node.right == null)
			{
				// When get leaf node And 
				// path contain all Odd nodes
				return 1;
			}
			return countOddNodePath(node.left) + 
              countOddNodePath(node.right);
		}
	}
	public static void main(String[] args)
	{
		// New of binary tree
		BinaryTree tree = new BinaryTree();
		/* 
		  Construct Binary Tree
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     / \    / \
		    7   6  1   3
		   / \    / \   
		  10  3  7   4
		*/
		// Add tree node
		tree.root = new TreeNode(5);
		tree.root.left = new TreeNode(3);
		tree.root.right = new TreeNode(9);
		tree.root.right.right = new TreeNode(3);
		tree.root.left.right = new TreeNode(6);
		tree.root.right.left = new TreeNode(1);
		tree.root.left.left = new TreeNode(7);
		tree.root.left.left.left = new TreeNode(10);
		tree.root.left.left.right = new TreeNode(3);
		tree.root.right.left.right = new TreeNode(4);
		tree.root.right.left.left = new TreeNode(7);
		/* 
		  Given Binary Tree
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     / \    / \
		    7   6  1   3
		   / \    / \   
		  10  3  7   4

		  Here
		  Odd path from root to leaf
		  -----------------------
		         5
		        /  \ 
		       /    \
		      3      9
		     /      / \
		    7      1   3
		     \    /     
		      3  7    
		  ---------------------
			5->3->7->3    
			5->9->1->7  
		    5->9->3  
		  ----------------------
		  Result : 3 [number of path]
		*/
        // Count odd nodes paths from root 
        // to leaf in a binary tree.
		System.out.println(tree.countOddNodePath(tree.root));
	}
}

Output

3

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