I was wondering if anybody could help me rework this method to find the height of a binary search tree. So far, my code looks like this. However, the answer I'm getting is larger than the actual height by 1. But when I remove the +1 from my return statements, it's less than the actual height by 1. I'm still trying to wrap my head around recursion with these BST. Any help would be much appreciated.
public int findHeight(){ if(this.isEmpty()){ return 0; } else{ TreeNode<T> node = root; return findHeight(node); } } private int findHeight(TreeNode<T> aNode){ int heightLeft = 0; int heightRight = 0; if(aNode.left!=null) heightLeft = findHeight(aNode.left); if(aNode.right!=null) heightRight = findHeight(aNode.right); if(heightLeft > heightRight){ return heightLeft+1; } else{ return heightRight+1; } }
The stick is held pointing straight up, at 90 degrees to your outstretched, straight arm. Carefully walk backwards until the top of the tree lines up with the top of your stick. Mark where your feet are. The distance between your feet and the tree is roughly equivalent to the height of the tree.
The height of a node is the number of edges from the node to the deepest leaf. The height of a tree is a height of the root. A full binary tree.is a binary tree in which each node has exactly zero or two children.
The problem lies in your base case.
"The height of a tree is the length of the path from the root to the deepest node in the tree. A (rooted) tree with only a node (the root) has a height of zero." - Wikipedia
If there is no node, you want to return -1 not 0. This is because you are adding 1 at the end.
So if there isn't a node, you return -1 which cancels out the +1.
int findHeight(TreeNode<T> aNode) { if (aNode == null) { return -1; } int lefth = findHeight(aNode.left); int righth = findHeight(aNode.right); if (lefth > righth) { return lefth + 1; } else { return righth + 1; } }
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