平衡二叉树（Balanced Binary Tree）具有以下性质：它是一棵空树或它的左右两个子树的高度差的绝对值不超过1。并且左右两个子树都是一棵平衡二叉树

（不是我们平时意义上的必须为搜索树）

判断一棵树是否为平衡二叉树：

 

可以暴力判断：每一颗树是否为平衡二叉树。

 

分析：

如果左右子树都已知是平衡二叉树，而左子树和右子树高度差绝对值不超过1，本树就是平衡的。

 

为此我们需要的信息：左右子树是否为平衡二叉树。左右子树的高度。

 

我们需要给父返回的信息就是：本棵树是否是平衡的、本棵树的高度。

 

定义结点和返回值：

	public static class Node {public int value;public Node left;public Node right;public Node(int data) {this.value = data;}}

	public static class ReturnType {public int level;   //深度public boolean isB;//本树是否平衡public ReturnType(int l, boolean is) {level = l;isB = is;}}

我们把代码写出来：

	// process(head, 1)public static ReturnType process(Node head, int level) {if (head == null) {return new ReturnType(level, true);}//取信息ReturnType leftSubTreeInfo = process(head.left, level + 1);if(!leftSubTreeInfo.isB) {return new ReturnType(level, false);     //左子树不是->返回}ReturnType rightSubTreeInfo = process(head.right, level + 1);if(!rightSubTreeInfo.isB) {return new ReturnType(level, false);     //右子树不是->返回}if (Math.abs(rightSubTreeInfo.level - leftSubTreeInfo.level) > 1) {return new ReturnType(level, false);     //左右高度差大于1->返回}return new ReturnType(Math.max(leftSubTreeInfo.level, rightSubTreeInfo.level), true);//返回高度和true（当前树是平衡的）}

我们不需要每次都返回高度，用一个全局变量记录即可。

对于其它二叉树问题，可能不止一个变量信息，所以，全局记录最好都养成定义数组的习惯。

下面贴出完整代码：

import java.util.LinkedList;
import java.util.Queue;public class Demo {public static class Node {public int value;public Node left;public Node right;public Node(int data) {this.value = data;}}public static boolean isBalance(Node head) {boolean[] res = new boolean[1];res[0] = true;getHeight(head, 1, res);return res[0];}public static class ReturnType {public int level;   //深度public boolean isB;//本树是否平衡public ReturnType(int l, boolean is) {level = l;isB = is;}}// process(head, 1)public static ReturnType process(Node head, int level) {if (head == null) {return new ReturnType(level, true);}//取信息ReturnType leftSubTreeInfo = process(head.left, level + 1);if(!leftSubTreeInfo.isB) {return new ReturnType(level, false);     //左子树不是->返回}ReturnType rightSubTreeInfo = process(head.right, level + 1);if(!rightSubTreeInfo.isB) {return new ReturnType(level, false);     //右子树不是->返回}if (Math.abs(rightSubTreeInfo.level - leftSubTreeInfo.level) > 1) {return new ReturnType(level, false);     //左右高度差大于1->返回}return new ReturnType(Math.max(leftSubTreeInfo.level, rightSubTreeInfo.level), true);//返回高度和true（当前树是平衡的}public static int getHeight(Node head, int level, boolean[] res) {if (head == null) {return level;//返回高度}//取信息//相同逻辑int lH = getHeight(head.left, level + 1, res);if (!res[0]) {return level;}int rH = getHeight(head.right, level + 1, res);if (!res[0]) {return level;}if (Math.abs(lH - rH) > 1) {res[0] = false;}return Math.max(lH, rH);//返回高度}public static void main(String[] args) {Node head = new Node(1);head.left = new Node(2);head.right = new Node(3);head.left.left = new Node(4);head.left.right = new Node(5);head.right.left = new Node(6);head.right.right = new Node(7);System.out.println(isBalance(head));}}