题目:
请实现一个函数来判断一棵二叉树是不是对称的。如果一棵二叉树和它的镜像一样,那么它是对称的。
实现
public class Solution59 {
public static boolean isSymmetrical2(BinaryTreeNode root) {
return isSymmetrical2(root, root);
}
private static boolean isSymmetrical2(BinaryTreeNode left, BinaryTreeNode right) {
if (left == null && right == null) {
return true;
}
if (left == null || right == null) {
return false;
}
if (left.value != right.value) {
return false;
}
return isSymmetrical2(left.left, right.right) && isSymmetrical2(left.right, right.left);
}
public static boolean isSymmetrical(BinaryTreeNode root){
if(root==null){
System.out.print("空树");
}
ArrayList node_list1 = new ArrayList();
//遍历树,将树各结点值存入list
traversal(node_list1,root);
//求树的镜像
mirrorTree(root);
ArrayList node_list2 = new ArrayList();
traversal(node_list2,root);
boolean result = compareList(node_list1,node_list2);
return result;
}
public static void traversal(ArrayList node_list,BinaryTreeNode root){
if(root==null){
return;
}
node_list.add(root.value);
traversal(node_list,root.left);
traversal(node_list,root.right);
}
public static void mirrorTree(BinaryTreeNode root){
if(root==null){
return;
}
BinaryTreeNode temp =root.left;
root.left = root.right;
root.right = temp;
mirrorTree(root.left);
mirrorTree(root.right);
}
public static boolean compareList(ArrayList list1,ArrayList list2){
if(list1.size()!=list2.size()){
return false;
}
int temp1=0;
int temp2=0;
for(int i=0;i<list1.size();i++){
temp1=(int)list1.get(i);
temp2=(int)list2.get(i);
if(temp1!=temp2){
return false;
}
}
return true;
}
public static void main(String[] args) {
test01();
test02();
}
private static void assemble(BinaryTreeNode node,
BinaryTreeNode left,
BinaryTreeNode right) {
node.left = left;
node.right = right;
}
// 1
// 2 2
// 4 6 6 4
// 8 9 10 11 11 10 9 8
public static void test01() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(2);
BinaryTreeNode n3 = new BinaryTreeNode(2);
BinaryTreeNode n4 = new BinaryTreeNode(4);
BinaryTreeNode n5 = new BinaryTreeNode(6);
BinaryTreeNode n6 = new BinaryTreeNode(6);
BinaryTreeNode n7 = new BinaryTreeNode(4);
BinaryTreeNode n8 = new BinaryTreeNode(8);
BinaryTreeNode n9 = new BinaryTreeNode(9);
BinaryTreeNode n10 = new BinaryTreeNode(10);
BinaryTreeNode n11 = new BinaryTreeNode(11);
BinaryTreeNode n12 = new BinaryTreeNode(11);
BinaryTreeNode n13 = new BinaryTreeNode(10);
BinaryTreeNode n14 = new BinaryTreeNode(9);
BinaryTreeNode n15 = new BinaryTreeNode(8);
assemble(n1, n2, n3);
assemble(n2, n4, n5);
assemble(n3, n6, n7);
assemble(n4, n8, n9);
assemble(n5, n10, n11);
assemble(n6, n12, n13);
assemble(n7, n14, n15);
assemble(n8, null, null);
assemble(n9, null, null);
assemble(n10, null, null);
assemble(n11, null, null);
assemble(n12, null, null);
assemble(n13, null, null);
assemble(n14, null, null);
assemble(n15, null, null);
System.out.println(isSymmetrical(n1));
}
// 1
// 2 2
// 4 5 6 4
// 8 9 10 11 11 10 9 8
public static void test02() {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(2);
BinaryTreeNode n3 = new BinaryTreeNode(2);
BinaryTreeNode n4 = new BinaryTreeNode(4);
BinaryTreeNode n5 = new BinaryTreeNode(5);
BinaryTreeNode n6 = new BinaryTreeNode(6);
BinaryTreeNode n7 = new BinaryTreeNode(4);
BinaryTreeNode n8 = new BinaryTreeNode(8);
BinaryTreeNode n9 = new BinaryTreeNode(9);
BinaryTreeNode n10 = new BinaryTreeNode(10);
BinaryTreeNode n11 = new BinaryTreeNode(11);
BinaryTreeNode n12 = new BinaryTreeNode(11);
BinaryTreeNode n13 = new BinaryTreeNode(10);
BinaryTreeNode n14 = new BinaryTreeNode(9);
BinaryTreeNode n15 = new BinaryTreeNode(8);
assemble(n1, n2, n3);
assemble(n2, n4, n5);
assemble(n3, n6, n7);
assemble(n4, n8, n9);
assemble(n5, n10, n11);
assemble(n6, n12, n13);
assemble(n7, n14, n15);
assemble(n8, null, null);
assemble(n9, null, null);
assemble(n10, null, null);
assemble(n11, null, null);
assemble(n12, null, null);
assemble(n13, null, null);
assemble(n14, null, null);
assemble(n15, null, null);
System.out.println(isSymmetrical(n1));
}
}