Subtree（java）

SubtreeShow result 

You have two every 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.

Have you met this question in a real interview? 

Yes

Example

T2 is a subtree of T1 in the following case:

       1                3      / \              / T1 = 2   3      T2 =  4        /       4

T2 isn't a subtree of T1 in the following case:

       1               3      / \               \T1 = 2   3       T2 =    4        /       4

package campus.lintCode;public class SubTree {public static void main(String[] args) {TreeNode treeNode1 = new TreeNode(1);TreeNode treeNode21 = new TreeNode(1);TreeNode treeNode22 = new TreeNode(1);TreeNode treeNode31 = new TreeNode(2);TreeNode treeNode32 = new TreeNode(3);TreeNode treeNode34 = new TreeNode(2);TreeNode treeNode43 = new TreeNode(4);TreeNode treeNode44 = new TreeNode(5);TreeNode treeNode45 = new TreeNode(3);treeNode1.left = treeNode21;treeNode1.right = treeNode22;treeNode21.left = treeNode31;treeNode21.right = treeNode32;treeNode22.right = treeNode34;treeNode32.left = treeNode43;treeNode32.right = treeNode44;treeNode34.left = treeNode45;TreeNode treeNode212 = new TreeNode(1);TreeNode treeNode312 = new TreeNode(2);TreeNode treeNode322 = new TreeNode(3);TreeNode treeNode432 = new TreeNode(4);TreeNode treeNode442 = new TreeNode(5);treeNode212.left = treeNode312;treeNode212.right = treeNode322;treeNode322.left = treeNode432;treeNode322.right = treeNode442;SubTree subTree = new SubTree();System.out.println(subTree.isSubtree(treeNode1, treeNode212));//System.out.println(subTree.priorSubtree(treeNode1,treeNode212));}    /**     * @param T1, T2: The roots of binary tree.     * @return: True if T2 is a subtree of T1, or false.     */    public boolean isSubtree(TreeNode T1, TreeNode T2) {        // write your code here         boolean result  = false;// 如何child=NULL,则说明child已经到达叶子了，无论parent是否到达一个叶子节点，都应该返回真if (T2 == null) {return true;}// 首先，如果程序执行到这里，则child一定不是NULL,如果parent=NULL,则说明parent不包含childif (T1 == null) {return false;}if (T1.val == T2.val) {// 当if条件满足的时候，则继续递归判断它们的左子树和右子树是否相等，只有两者相等才返回真 result = isSubtreeDef(T1,T2);} if(!result){result= (isSubtree(T1.left, T2) || isSubtree(T1.right, T2));}return result;}public boolean isSubtreeDef(TreeNode T1, TreeNode T2) {// write your code here// 如何child=NULL,则说明child已经到达叶子了，无论parent是否到达一个叶子节点，都应该返回真if (T2 == null&&T1 == null) {return true;}if (T1!=null&&T2!=null&&T1.val == T2.val) {// 当if条件满足的时候，则继续递归判断它们的左子树和右子树是否相等，只有两者相等才返回真return (isSubtreeDef(T1.left, T2.left) && isSubtreeDef(T1.right, T2.right));} return false;}}class TreeNode {public int val;public TreeNode left, right;public TreeNode(int val) {this.val = val;this.left = this.right = null;}}