【剑指Offer】17、树的子结构

题目描述

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

题解：递归

 1 public static boolean HasSubtree(TreeNode root1,TreeNode root2) {
 2         if(root1==null||root2==null){
 3             return false;
 4         }
 5        return tree1HasTree2(root1,root2)||HasSubtree(root1.left,root2)||HasSubtree(root1.right,root2);
 6     }
 7     public static boolean tree1HasTree2(TreeNode root1,TreeNode root2){
 8         if(root1==null){
 9             return false;
10         }
11         if(root2==null){
12             return true;
13         }
14         return root1.val==root2.val&&tree1HasTree2(root1.left,root2.left)
15             &&tree1HasTree2(root1.right,root2.right);
16     }

初始化树：

 1 public static class TreeNode {
 2         int val = 0;
 3         TreeNode left = null;
 4         TreeNode right = null;
 5         public TreeNode(int val) {
 6             this.val = val;
 7         }
 8     }
 9  private static List<TreeNode> nodeList = null;
10     public static TreeNode createBinTree(int[] array) {
11         nodeList=new LinkedList<TreeNode>();
12         // 将一个数组的值依次转换为TreeNode节点
13         for (int nodeIndex = 0; nodeIndex < array.length; nodeIndex++) {
14             nodeList.add(new TreeNode(array[nodeIndex]));
15         }
16         // 对前lastParentIndex-1个父节点按照父节点与孩子节点的数字关系建立二叉树
17         for (int parentIndex = 0; parentIndex < array.length / 2 - 1; parentIndex++) {
18             // 左孩子
19             nodeList.get(parentIndex).left = nodeList.get(parentIndex * 2 + 1);
20             // 右孩子
21             nodeList.get(parentIndex).right = nodeList.get(parentIndex * 2 + 2);
22         }
23         // 最后一个父节点:因为最后一个父节点可能没有右孩子，所以单独拿出来处理
24         int lastParentIndex = array.length / 2 - 1;
25         // 左孩子
26         nodeList.get(lastParentIndex).left = nodeList.get(lastParentIndex * 2 + 1);
27         // 右孩子,如果数组的长度为奇数才建立右孩子
28         if (array.length % 2 == 1) {
29             nodeList.get(lastParentIndex).right = nodeList.get(lastParentIndex * 2 + 2);
30         }
31         return nodeList.get(0);
32     }

测试：

1 public static void main(String[] args) {
2         int[] tree1={8,8,7,9,2,-1,-1,-1,-1,4,7};
3         int[] tree2={8,9,2};
4         TreeNode root1 = createBinTree(tree1);
5         TreeNode root2 = createBinTree(tree2);
6         boolean hasSubtree = HasSubtree(root1, root2);
7         System.out.println(hasSubtree);
8     }
9 输出：true

标签：TreeNode,17,Offer,int,get,子结构,nodeList,root1,root2
来源： https://www.cnblogs.com/Blog-cpc/p/12369361.html