572. Subtree of Another Tree

Problem Description

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:

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

Example 2:

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

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]
• -10⁴ <= root.val <= 10⁴
• -10⁴ <= subRoot.val <= 10⁴

Python Solution

# 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:
        if not root:
            return False
        
        if self.isSameTree(root, subRoot):
            return True
        
        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)
    
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        if not p and not q:
            return True
        if not p or not q:
            return False
        
        return (p.val == q.val and 
                self.isSameTree(p.left, q.left) and 
                self.isSameTree(p.right, q.right))

Java Solution

/**
 * 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) {
        if (root == null) {
            return false;
        }
        
        if (isSameTree(root, subRoot)) {
            return true;
        }
        
        return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
    }
    
    private boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        }
        if (p == null || q == null) {
            return false;
        }
        
        return p.val == q.val && 
               isSameTree(p.left, q.left) && 
               isSameTree(p.right, q.right);
    }
}

C++ Solution

/**
 * 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) {
        if (!root) {
            return false;
        }
        
        if (isSameTree(root, subRoot)) {
            return true;
        }
        
        return isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot);
    }
    
private:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if (!p && !q) {
            return true;
        }
        if (!p || !q) {
            return false;
        }
        
        return p->val == q->val && 
               isSameTree(p->left, q->left) && 
               isSameTree(p->right, q->right);
    }
};

JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} subRoot
 * @return {boolean}
 */
var isSubtree = function(root, subRoot) {
    if (!root) {
        return false;
    }
    
    if (isSameTree(root, subRoot)) {
        return true;
    }
    
    return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
};

function isSameTree(p, q) {
    if (!p && !q) {
        return true;
    }
    if (!p || !q) {
        return false;
    }
    
    return p.val === q.val && 
           isSameTree(p.left, q.left) && 
           isSameTree(p.right, q.right);
}

C# Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public bool IsSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null) {
            return false;
        }
        
        if (IsSameTree(root, subRoot)) {
            return true;
        }
        
        return IsSubtree(root.left, subRoot) || IsSubtree(root.right, subRoot);
    }
    
    private bool IsSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) {
            return true;
        }
        if (p == null || q == null) {
            return false;
        }
        
        return p.val == q.val && 
               IsSameTree(p.left, q.left) && 
               IsSameTree(p.right, q.right);
    }
}