Subtree

Subtree (lintcode)

Description
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.

Notice
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.

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

解题思路

使用递归的思路，首先判断T2和T1是否相等，如果不相等则接着让T2和T1左子树、右子树比较。这里面非常重要的一点是判断最基本的情况。

Java 实现

/**
 * 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
;
        }


if
 (isEqual(T1, T2)) {

return
true
;
        }


// if the tree with root as current node doesn't match then
// try left and right subtrees one by one
return
 isSubtree(T1.left, T2) || isSubtree(T1.right, T2);
    }


private
boolean
isEqual
(TreeNode root1, TreeNode root2)
{

if
 (root1 == 
null
&
&
 root2 == 
null
) {

return
true
;
        }

if
 (root1 == 
null
 || root2 == 
null
) {

return
false
;
        }


// check if the data of both roots is same and 
// data of left and right subtrees are also same
return
 root1.val == root2.val 
&
&

                isEqual(root1.left, root2.left) 
&
&

                isEqual(root1.right, root2.right);
    }

}