cpp-flashcards/org/study_deck_02/dsa/trees/0572-subtree-of-another-tree.org

#+ANKI_DECK: study_deck_02
* TODO 0572. Subtree of Another Tree :easy:
:PROPERTIES:
:NEETCODE: [[file:../../roadmap.org::*0572. Subtree of Another Tree][0572. Subtree of Another Tree]]
:END:

Given the roots of two binary trees ~root~ and ~subRoot~, return ~true~ if there is a subtree of ~root~ with the same structure and node values of~ subRoot~ and ~false~ otherwise.

A subtree of a binary tree ~tree~ is a tree that consists of a node in ~tree~ and all of this node's descendants. The tree ~tree~ could also be considered as a subtree of itself.

*Example 1:*


#+begin_src
Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true
#+end_src


*Example 2:*


#+begin_src
Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
Output: false
#+end_src


*Constraints:*

 - The number of nodes in the ~root~ tree is in the range ~[1, 2000]~.

 - The number of nodes in the ~subRoot~ tree is in the range ~[1, 1000]~.

 - ~-10^{4} <= root.val <= 10^{4}~

 - ~-10^{4} <= subRoot.val <= 10^{4}~

** TODO Approach
Write your approach here.

** TODO Python
#+begin_src python :lc-problem 572 :lc-lang python3
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
#+end_src

** TODO C++
#+begin_src cpp :lc-problem 572
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        
    }
};
#+end_src