Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    / \
   4   5
  / \
 1   2
Given tree t:
   4 
  / \
 1   2
Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

     3
    / \
   4   5
  / \
 1   2
    /
   0
Given tree t:
   4
  / \
 1   2
Return false.

Solution1：Tree 遍歷

思路：
Time Complexity: O(N^2) Space Complexity: O(N) 遞歸緩存

Solution2：轉(序列化) 先序 + 中序，根據

思路：轉 先序 + 中序后，分別根據先序和中序是否contains判斷，因為先序 + 中序可以唯一確定一個樹。contains時如果使用的是KMP，就是O(N)
Time Complexity: O(N) Space Complexity: O(N)

Solution1 Code：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        if (s == null) return false;
        if (isSame(s, t)) return true;
        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    private boolean isSame(TreeNode s, TreeNode t) {
        if (s == null && t == null) return true;
        if (s == null || t == null) return false;
        
        if (s.val != t.val) return false;
        
        return isSame(s.left, t.left) && isSame(s.right, t.right);
    }
}

Solution2 Code：

public class Solution {
 public boolean isSubtree(TreeNode s, TreeNode t) {
        String spreorder = generatepreorderString(s); 
        String tpreorder = generatepreorderString(t);
        
        return spreorder.contains(tpreorder) ;
    }
    public String generatepreorderString(TreeNode s){
        StringBuilder sb = new StringBuilder();
        Stack<TreeNode> stacktree = new Stack();
        stacktree.push(s);
        while(!stacktree.isEmpty()){
           TreeNode popelem = stacktree.pop();
           if(popelem==null)
              sb.append(",#"); // Appending # inorder to handle same values but not subtree cases
           else      
              sb.append(","+popelem.val);
           if(popelem!=null){
                stacktree.push(popelem.right);    
                stacktree.push(popelem.left);  
           }
        }
        return sb.toString();
    }
}