Counting left-child nodes in a Tree
I am supposed to implement a recursive method that counts the amount of left-child tree nodes. My code so far is:
private int countLeftNodes(IntTreeNode node){
int c = 0;
if (node != null){
c = 1 + countLeftNodes(node.left);
countLeftNodes(node.right);
}
return c;
}
It returns a number much smaller than what it should be. I have a feeling that my traversal is off because it seems to only count the very left child nodes, and then terminates. When I call this method on an IntTree of size 16 I should get 8 left-child nodes, 7 right-child nodes, and one root, but inst开发者_Python百科ead I get 4 left-child nodes.
You never count the left nodes in the right tree.
private int countLeftNodes(IntTreeNode node)
{
int c = 0;
if (node.left != null)
{
c += 1 + countLeftNodes(node.left);
}
if(node.right != null)
{
c += countLeftNodes(node.right);
}
return c;
}
To count left-child nodes you can do:
private int countLeftNodes(IntTreeNode node) {
// no tree no left-child nodes
if(node == null) {
return 0;
}
// left-child count of current node.
int c = 0;
// does the current node have a left-child ?
if (node.left != null){
c = 1;
}
// return left-child count of current node +
// left-child count of left and right subtrees
return c + countLeftNodes(node.left) + countLeftNodes(node.right);
}
private int countLeftNodes(IntTreeNode node){
int c = 0;
if (node != null){
if(node.left!=null) {
c = 1 + countLeftNodes(node.left);
}
if(node.right!=null){
c +=countLeftNodes(node.right);
}
}
return c;
}
easiest place to check that is in the parent.
private int countLeftNodes(IntTreeNode node){
int c = 0;
if(node.left != null)
{
c++;
c+= countLeftNodes(node.left)
}
if(node.right != null)
{
c+= countLeftNodes(node.right);
}
return c;
}
My favorite style when using recursion is to use a wrapper function of some sort where the main method calls another that does the grunt work:
private int countLeftNodes(IntTreeNode node){
int totalCount = reallyCountLeftNodes(IntTreeNode node, 0);
return totalCount;
}
private int reallyCountLeftNodes(IntTreeNode n, Int sum){
if (n.left == NULL && n.right == NULL){ //check if we've hit rock bottom
return sum;
} else if (n.left == NULL) { //if the node's left is nil, go right
reallyCountLeftNodes(n.right, sum++);
} else {
reallyCountLeftNodes(n.left, sum++); // Going as far left as possible!
}
}
Notice how the main function calls another. I find this style to be cleaner and easier to understand. Also, the second function has a count variable for you to use.
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