Tree_Graph 判断T2是否为T1的子树 @CareerCup

来源：互联网发布：剑雨江湖天罡进阶数据编辑：程序博客网时间：2024/04/30 09:26

原文：

You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1

译文：

有两棵很大的二叉树：T1有上百万个结点，T2有上百个结点。写程序判断T2是否为T1的子树。

思路：

1）因为中序前序遍历就可以决定一棵树，把两棵树的中序前序都写出来，如果两个遍历中T2都是T1的子串，可以确定T2是T1的子树。

当然这会遇到无法区分

的情况。因为

T1, in order: 3,3 T1, pre order 3,3

T2, in order: 3,3 T2, pre order 3,3

所以必须也算入null节点，所以

T1, in order: 0, 3, 0, 3, 0 T1, pre order 3, 3, 0, 0, 0

T2, in order: 0, 3, 0, 3, 0 T2, pre order 3, 0, 3, 0, 0

这种方法时间复杂度：O(n+m), 空间复杂度：O(n+m) , n,m分别为两个树的节点数。当海量数据时不适用！

2）递归！从根节点往下递归，判断子树是否存在其的左子树或右子树内。

时间复杂度：O(nm), 空间复杂度：O(log(n) + log(m))

但实际上，在真实的运行时，递归假如发现满足子树的条件就直接返回，而不会继续查找。因此递归法在实际中无论时间空间更优！

package Tree_Graph;import CtCILibrary.AssortedMethods;import CtCILibrary.TreeNode;public class S4_8 {// 判断t2是否是t1的子树public static boolean isSubTree(TreeNode t1, TreeNode t2) {if(t2 == null) {// 空树始终是另一个树的子树return true;}if(t1 == null) {// 此时r2非空，非空树不可能是一个空树的子树return false;}if(t1.data == t2.data) {// 找到两树匹配if(isSameTree(t1, t2)){return true;}}// 继续在r1的左子树和右子树里找匹配return isSubTree(t1.left, t2) || isSubTree(t1.right, t2);}// 判断两棵树是否相同public static boolean isSameTree(TreeNode t1, TreeNode t2) {if(t1==null && t2==null) {return true;}if(t1==null || t2==null) {return false;}if(t1.data != t2.data) {return false;}return isSameTree(t1.left, t2.left) && isSameTree(t1.right, t2.right);}public static void main(String[] args) {// t2 is a subtree of t1int[] array1 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};int[] array2 = {2, 4, 5, 8, 9, 10, 11};TreeNode t1 = AssortedMethods.createTreeFromArray(array1);TreeNode t2 = AssortedMethods.createTreeFromArray(array2);if (isSubTree(t1, t2))System.out.println("t2 is a subtree of t1");elseSystem.out.println("t2 is not a subtree of t1");// t4 is not a subtree of t3int[] array3 = {1, 2, 3};TreeNode t3 = AssortedMethods.createTreeFromArray(array1);TreeNode t4 = AssortedMethods.createTreeFromArray(array3);if (isSubTree(t3, t4))System.out.println("t4 is a subtree of t3");elseSystem.out.println("t4 is not a subtree of t3");}}

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