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);
    }
}