lintcode-easy-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.

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


/**
 * 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(T1 == null && T2 == null)
            return true;
        if(T1 == null)
            return false;
        if(T2 == null)
            return true;
        
        if(isSame(T1, T2))
            return true;
        else
            return isSubtree(T1.left, T2) || isSubtree(T1.right, T2);
        
    }
    
    public 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);
    }
}

 

posted @ 2016-03-07 08:15  哥布林工程师  阅读(129)  评论(0编辑  收藏  举报