LintCode Subtree

原题链接在这里：http://www.lintcode.com/en/problem/subtree/

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 ofT1.

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

Note

A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.

Time Complexity: O(m*n), m是T1的node数, n 是T2的node 数. Space: O(logm). isSame用logn, isSubtree用logm.

AC Java:

 /**
  * Definition of TreeNode:
  * public class TreeNode {
  *     public int val;
  *     public TreeNode left, right;
  *     public TreeNode(int val) {
  *         this.val = val;
  *         this.left = this.right = null;
  *     }
  * }
  */
 public class Solution {
     /**
      * @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
         if(T2 == null){
             return true;
         }
         if(T1 == null){
             return false;
         }
         return isSame(T1,T2) || isSubtree(T1.left, T2) || isSubtree(T1.right, T2);
     }

     private boolean isSame(TreeNode T1, TreeNode T2){
         if(T1 == null && T2 == null){
             return true;
         }
         if(T1 == null || T2 == null){
             return false;
         }
         if(T1.val != T2.val){
             return false;
         }
         return isSame(T1.left, T2.left) && isSame(T1.right, T2.right);
     }
 }