572. Subtree of Another Tree

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.

 

//Approach1: recursive: 思路就是对于s树的每一个点和t进行isSameTree判断
//Time: O(n!), Space: O(h)
 
   public boolean isSubtree(TreeNode s, TreeNode t) {
        if (s == null) return false;//切勿忘记这一句，否则下面s.left和s.right就溢出
        
        if (isSameTree(s, t)) {
            return true;
        }
        
        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }
    
    private boolean isSameTree(TreeNode s, TreeNode t) {
        if (s == null && t == null) {
            return true;
        }
        
        if (s == null || t == null) {
            return false;
        }
        
        if (s.val == t.val) {
            return isSameTree(s.left, t.left) && isSameTree(s.right, t.right);
        }
        
        return false;
    }

//Approach2: traversal: 思路就是分别遍历s和t，看t是不是s的子集
//Time: O(n), Space: O(n)
    public boolean isSubtree(TreeNode s, TreeNode t) {
        StringBuilder s1 = serilize(s, new StringBuilder());
        StringBuilder t1 = serilize(t, new StringBuilder());
        return s1.toString().contains(t1.toString());
    }
    
    private StringBuilder serilize(TreeNode root, StringBuilder sb) {
        if (root == null) {
            sb.append(",#");//注意这里很重要，要先加，再加#，否则12和2就被判定为相同
            return sb;
        }
        
        sb.append("," + root.val);
        serilize(root.left, sb);
        serilize(root.right, sb);
        return sb;
    }