Subtree of Another Tree

Practice here

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.

Approach#

• Intuition: At every node in the main tree you have a candidate root that could match subRoot. If the subtree rooted there is structurally and value-identical to subRoot, return true; otherwise recurse into both children to keep looking.
• Mechanics: A helper checkIdentical performs a simultaneous DFS on both trees, returning true only when both pointers are null together, both non-null with equal values, and their left and right subtrees also match recursively. The outer isSubtree calls this helper at each node, short-circuiting as soon as a match is found.
• Trade-off: This is the natural recursive solution — simple to reason about and correct. The worst-case O(n × m) cost occurs on degenerate trees (e.g., a linked list); for balanced trees the average cost is much lower. A serialization-based approach can reduce this to O(n + m) at the expense of extra space and edge-case handling.

/**
 * 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 checkIdentical(TreeNode* root1, TreeNode* root2)
    {
        if(!root1 && !root2)
            return true;
        if(!root1 || !root2)
            return false;

        return root1->val == root2->val && checkIdentical(root1->left, root2->left) && checkIdentical(root1->right, root2->right);
    }

    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        if(!root)
            return false;

        if(checkIdentical(root, subRoot))
            return true;

        return isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot);
    }
};

Time Complexity: O(n × m),

• In the worst case, for every node in the main tree (n nodes), you may need to call checkIdentical and compare it with the subRoot (which takes up to m time).

Space Complexity: O(h)

• You are using recursive DFS in both isSubtree() and checkIdentical(), so the maximum depth of the recursion stack is the height of the tree → h.