Subtree of Another Tree (Easy Facebook)

update May 11,2018 21:58

LeetCode

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.

Basic Idea:

写一个 isSame 的 helper function，判断两个node所代表的是否是同样的树，然后递归调用。两层递归嵌套。

• C++ Code：

  class Solution {
    bool isSame(TreeNode* node1, TreeNode* node2) {
        if (node1 == nullptr && node2 == nullptr) return true;
        else if (node1 == nullptr || node2 == nullptr) return false;
        else if (node1->val != node2->val) return false;
        bool left = isSame(node1->left, node2->left);
        bool right = isSame(node1->right, node2->right);
        return left && right;
    }
  public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if (t == nullptr) return true;
        else if (s == nullptr) return false;
        if (s->val == t->val && isSame(s, t)) return true;
        else return isSubtree(s->left, t) || isSubtree(s->right, t);
    }
  };
  

update 2018-10-07 21:48:46

Update：Simpler java code

• Java Code:

  更加简洁的写法，直接对比s的每个子树和t是否相等，不需要提前判断值是否相同。

  class Solution {
      public boolean isSubtree(TreeNode s, TreeNode t) {
          if (s == null) return false;
          return helper(s, t) || isSubtree(s.left, t) || isSubtree(s.right, t);
      }
  
      private boolean helper(TreeNode s, TreeNode t) {
          if (s == null && t == null) return true;
          else if (s == null || t == null) return false;
          else if (s.val != t.val) return false;
          return helper(s.left, t.left) && helper(s.right, t.right);
      }
  }