LintCode 93. Balanced Binary Tree 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/balanced-binary-tree/
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example
Given binary tree A = {3,9,20,#,#,15,7}, B = {3,#,20,15,7}
A) 3 B) 3
/ \ \
9 20 20
/ \ / \
15 7 15 7
The binary tree A is a height-balanced binary tree, but B is not.
解题思路
题目要求去判断一颗二叉树是否高度平衡，即任何左右两树高度差不得大于1。
实现可以参考下面代码，先构造一个ResultType的类别，包括最大高度和是否平衡的属性。
接着建一个helper方法，分支返回对树支点进行判断后的ResultType。
参考代码
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: True if this Binary tree is Balanced, or false.
*/
private class ResultType {
private int maxHeight;
private boolean isBalanced;
public ResultType(int maxHeight, boolean isBalanced) {
this.maxHeight = maxHeight;
this.isBalanced = isBalanced;
}
}
public boolean isBalanced(TreeNode root) {
return helper(root).isBalanced;
}
private ResultType helper(TreeNode root) {
if (root == null) {
return new ResultType(0, true);
}
ResultType left = helper(root.left);
ResultType right = helper(root.right);
if (!left.isBalanced || !right.isBalanced) {
return new ResultType(Integer.MIN_VALUE, false);
}
if (Math.abs(left.maxHeight - right.maxHeight) > 1) {
return new ResultType(Integer.MIN_VALUE, false);
}
return new ResultType(Math.max(left.maxHeight, right.maxHeight) + 1, true);
}
}