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：递归判断

思路：
Time Complexity: O(m*n) Space Complexity: O(m+n) 递归缓存

Solution2：先序列化tree

思路：先序列化tree，再判断持否contains，如果判断contains方式用kmp，则可线性时间
Time Complexity: O(mn) Space Complexity: O(m+n)
如KMP:
Time Complexity: O(m+n) Space Complexity: O(m+n)

Solution1 Code：

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();
    }
}