剑指offer36--二叉树的深度

题目一：输入一棵二叉树的根结点，求该树的深度。从根结点到叶子点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。

二、平衡二叉树

输入一棵二叉树的根结点，判断该树是不是平衡二叉树。如果某二叉树中任意结点的左右子树的深度相差不超过1 ，那么它就是一棵平衡二叉树。

三、思想

四、程序

package 剑指offer;
/*题目一：输入一棵二叉树的根结点，求该树的深度。从根结点到叶子点依次经过的结点
*（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。
*/
public class Test39 {
public static int TreeDeepth(BinaryNode root){
if(root == null){
return 0;
}

int left = TreeDeepth(root.left);
int right = TreeDeepth(root.right);

if(left > right){
return left + 1;
}else{
return right + 1;
}
}

// 判断一个数是否是平衡二叉树
public static boolean isBalanced(BinaryNode root){
if(root == null){
return true;
}

int left = TreeDeepth(root.left);
int right = TreeDeepth(root.right);

int differ = left - right;
if(differ > 1 || differ < -1){
return false;
}

return isBalanced(root.left) && isBalanced(root.right);
}

public static boolean isBalanced2(BinaryNode root){
if(root == null){
return true;
}

int left = TreeDeepth(root.left);
int right = TreeDeepth(root.right);

int differ = left - right;
if(differ > 1 || differ < -1){
return false;
}

return isBalanced(root.left) && isBalanced(root.right);
}

public static void main(String args[]){
// 完全二叉树
//             1
//         /      \
//        2        3
//       /\       / \
//      4  5     6   7
BinaryNode n1 = new BinaryNode(1);
BinaryNode n2 = new BinaryNode(1);
BinaryNode n3 = new BinaryNode(1);
BinaryNode n4 = new BinaryNode(1);
BinaryNode n5 = new BinaryNode(1);
BinaryNode n6 = new BinaryNode(1);
BinaryNode n7 = new BinaryNode(1);

n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n3.left = n6;
n3.right = n7;

System.out.println(isBalanced(n1));
//System.out.println(isBalanced2(n1));
System.out.println("----------------");
}
}
class BinaryNode{
int value;
BinaryNode left;
BinaryNode right;

public BinaryNode(int value){
this.value = value;
}
}