Created
July 1, 2014 16:11
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[CC150][4.1/4th e.d] Implement a function to check if a tree is balanced For the purposes of this question, a balanced tree is defined to be a tree such that no two leaf nodes differ in distance from the root by more than one.
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public class BST<Key extends Comparable<Key>> { | |
private Node root; | |
private class Node { | |
private Key key; | |
private Node left, right; | |
public Node(Key key) | |
{ this.key = key; } | |
} | |
public BST(Key[] keys) { | |
for (Key key : keys) | |
put(key); | |
} | |
public void put(Key key) { | |
root = put(root, key); | |
} | |
private Node put(Node x, Key key) { | |
if (x == null) return new Node(key); | |
int cmp = key.compareTo(x.key); | |
if (cmp < 0) x.left = put(x.left, key); | |
else if (cmp > 0) x.right = put(x.right, key); | |
return x; | |
} | |
public boolean isBalanced() { | |
if (root == null) return true; | |
int[] lv = countLv(root); | |
System.out.println(lv[0]+" "+lv[1]); | |
if ((lv[0] - lv[1]) >= 2) return false; | |
return true; | |
} | |
private int[] countLv(Node x) { | |
int[] result = {0,0}; | |
if (x == null) | |
return result; | |
int[] countNextLeft = countLv(x.left); | |
int[] countNextRight = countLv(x.right); | |
result[0] = Math.max(countNextLeft[0], countNextRight[0]) + 1; | |
result[1] = Math.min(countNextLeft[1], countNextRight[1]) + 1; | |
return result; | |
} | |
public static void main(String[] args) { | |
// TODO Auto-generated method stub | |
Integer data[] = {7,4,9,2,5,8,10,1,3,6,11,0}; | |
/* 7 | |
* / \ | |
* / \ | |
* / \ | |
* 4 9 | |
* / \ / \ | |
* 2 5 8 10 | |
* / \ \ \ | |
* 1 3 6 11 | |
* / | |
* 0 | |
*/ | |
BST<Integer> test = new BST<Integer>(data); | |
if (test.isBalanced()) | |
System.out.println("Balanced"); | |
else System.out.println("Not Balanced"); | |
} | |
} |
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