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.

Accepted

127,861

Submissions

300,452

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if (s == NULL) {
            return t == NULL;
        }
        
        if (s -> val == t -> val) {
            bool ans = check(s, t);
            if (ans) {
                return true;
            }
        }
        return isSubtree(s -> left, t) || isSubtree(s -> right, t);
    }
    
    bool check(TreeNode* s, TreeNode* t) {
        if (s == NULL) {
            return t == NULL;
        }
        if (t == NULL) {
            return s == NULL;
        }
        if (s -> left == NULL && s -> right == NULL && t -> left == NULL && t -> right == NULL) {
            return s -> val == t -> val;
        } else {
            return (s -> val == t -> val) && check(s -> left, t -> left) && check(s -> right, t -> right);
        }
        return false;
    }
};


========= Cleaner one =======

bool isSubtree(TreeNode* s, TreeNode* t) { if(!s) return false; if (isSame(s,t)) return true; return isSubtree(s->left,t) || isSubtree(s->right,t); } bool isSame(TreeNode *s, TreeNode *t) { if (!s && !t) return true; if (!s || !t) return false; if (s->val != t->val) return false; return isSame(s->left, t->left) && isSame(s->right, t->right); }