0572 - Subtree of Another Tree (Easy)

Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        boolean contains = Objects.nonNull(subRoot) ? containsTree(root, subRoot) : true;
        return contains;
    }

    private boolean containsTree(TreeNode t1, TreeNode t2) {
        // Big tree t1 is empty, not a matching case
        if (Objects.isNull(t1)) return false;
        if (Objects.equals(t1.val, t2.val) && matchTree(t1, t2)) return true;
        return containsTree(t1.left, t2) || containsTree(t1.right, t2);
    }

    private boolean matchTree(TreeNode t1, TreeNode t2) {
        // nothing left in the subtree to compare
        if (Objects.isNull(t1) && Objects.isNull(t2)) return true;
        // one tree is empty, therefore trees don't match
        if (Objects.isNull(t1) || Objects.isNull(t2)) return false;
        // value doesn't match
        if (!Objects.equals(t1.val, t2.val)) return false;
        // Repeat until the last node of both left and right
        return matchTree(t1.left, t2.left) && matchTree(t1.right, t2.right);
    }
}

# 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:
    # Time O(mn) - for every node in tree, n, we may have to check it against subRoot, m.
    # Space: O(hm+hn) - height of m, subRoot and height of n, root.
    # isSubtree is DFS solution which with our stack holding the current path
    # which will scale with hn, and the isSameTree function will hold up to hm
    # calls inside the call stack as it scales with the height of subRoot.
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        # check p and q exists and the values are equal.
        if p and q and p.val == q.val:
            # are equal, recursively check corresponding left and right children.
            return self.isSameTree(p.left, q.left) and self.isSameTree(p.right,q.right)
        # aren't equal. p is q handles both are None -> return True else 1 is None
        # and the other a node -> return False
        return p is q

    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
        # initialize stack with our root
        stack = [root]
        # traverse the root tree.
        while stack:
            # pop current node off.
            node = stack.pop()
            # node matches subRoot root node, check same tree.
            if node.val == subRoot.val:
                # check if same tree.
                if self.isSameTree(node, subRoot):
                    # they are, we can return True
                    return True
            # add left subtree child node to the stack
            if node.left:
                stack.append(node.left)
            # add right subtree child node to the stack
            if node.right:
                stack.append(node.right)
        # made it through without finding subtree inside, return False.
        return False