Subtree 245

Question

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

Solution

非递归版本：非递归版本用BFS遍历T1,找所有与T2值相等的点，作为备选节点。然后比较以这些节点为根的子树是否和T2等同。

递归版本：先看T1和T2是否等同，若不等同，则递归看T1的左右子树是否和T2等同。

代码如下：

/**
 * 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;
        }

        //非递归 Version
        // Queue<TreeNode> queue = new LinkedList<TreeNode>();
        // queue.offer(T1);

        // TreeNode root = new TreeNode(T2.val + 1);
        // ArrayList<TreeNode> candidates = new ArrayList<TreeNode>();

        // while(!queue.isEmpty()){
        //     TreeNode curt = queue.poll();
        //     if(curt.val == T2.val){
        //         root = curt;
        //         candidates.add(curt);
        //     }
        //     if(curt.left != null){
        //         queue.offer(curt.left);
        //     }
        //     if(curt.right != null){
        //         queue.offer(curt.right);
        //     }
        // }

        // if(root.val != T2.val){
        //     return false;
        // }

        // for(TreeNode node : candidates){
        //     if(isIdentical(node, T2)){
        //         return true;
        //     }
        // }

        // return false;

        //递归 Version
        if(isIdentical(T1, T2)){
            return true;
        }

        if(isSubtree(T1.left, T2) || isSubtree(T1.right, T2)){
            return true;
        }

        return false;
    }

    private boolean isIdentical(TreeNode a, TreeNode b){
        if(a == null || b == null){
            return a == b;
        }

        if(a.val != b.val){
            return false;
        }

        return isIdentical(a.left, b.left) && isIdentical(a.right, b.right);
    }
}