剑指：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。

我们规定空树不是任何树的子结构。

样例

树 A：

 8
 / \
8 7
 / \
9 2
 / \
4 7

树 B：

 8
 / \
9 2

返回 true ,因为 B 是 A 的子结构。

 

解法
递归方式遍历：

• 在树 A 中找到和树 B 的根结点值一样的结点 R；
• 判断树 A 以 R 为根结点的子树是否包含与树 B 一样的结构。

 

package demo;public class Solution {//Definition for a binary tree node.public static class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) { val = x; }}public static boolean hasSubTree(TreeNode pRoot1, TreeNode pRoot2){boolean res = false;if(pRoot1 != null && pRoot2 != null){if(pRoot1.val == pRoot2.val){res = isSame(pRoot1, pRoot2);}if(!res){//res==falseres = hasSubTree(pRoot1.left, pRoot2);}if(!res){res = hasSubTree(pRoot1.right, pRoot2);}}return res;}private static boolean isSame(TreeNode root1, TreeNode root2) {if(root2==null){return true;}if(root1==null || root1.val !=root2.val){return false;}return isSame(root1.left, root2.left) && isSame(root1.right, root2.right);}public static void main(String[] args) {TreeNode R1 = new TreeNode(8);  TreeNode n2 = new TreeNode(8); R1.left = n2;  TreeNode n3 = new TreeNode(7); R1.right = n3;TreeNode n4 = new TreeNode(9); n2.left = n4;TreeNode n5 = new TreeNode(2); n2.right = n5;TreeNode n6 = new TreeNode(4); n5.left = n6;TreeNode n7 = new TreeNode(7); n5.right = n7;TreeNode R2 = new TreeNode(8); TreeNode ns2 = new TreeNode(9); R2.left=ns2;TreeNode ns3 = new TreeNode(2); R2.right=ns3;boolean t = hasSubTree(R1, R2);System.out.println(t);}
}