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IsTreeBalanced.java
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IsTreeBalanced.java
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/*
* Copyright (C) 2014 Pedro Vicente Gómez Sánchez.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package com.github.pedrovgs.problem65;
import com.github.pedrovgs.binarytree.BinaryNode;
/**
* Implement a function to check if a binary tree is balanced. For the purposes of this question, a
* balanced tree is defined to be a tree such that the heights of the two subtrees of any node
* never differ by more than one.
*
* @author Pedro Vicente Gómez Sánchez.
*/
public class IsTreeBalanced {
/**
* Recursive algorithm to check if a binary tree is balanced. The complexity order in time terms
* of this algorithm is equals to O(N) where N is the number of nodes in the binary tree. In
* space terms, the complexity order is O(1) because we are not using any auxiliary data
* structure to resolve this problem. This algorithm is based on the tree height because the
* balanced tree definition is based on this.
*/
public boolean check(BinaryNode root) {
if (root == null) {
return true;
} else {
int heightDifference = getRootHeight(root.getLeft()) - getRootHeight(root.getRight());
return Math.abs(heightDifference) <= 1;
}
}
private int getRootHeight(BinaryNode tree) {
if (tree == null) {
return 0;
} else {
return 1 + Math.max(getRootHeight(tree.getLeft()), getRootHeight(tree.getRight()));
}
}
}