572. Subtree of Another Tree

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

Solutions

Solution 1: DFS

We define a helper function $\textit{same}(p, q)$ to determine whether the tree rooted at $p$ and the tree rooted at $q$ are identical. If the root values of the two trees are equal, and their left and right subtrees are also respectively equal, then the two trees are identical.

In the $\textit{isSubtree}(\textit{root}, \textit{subRoot})$ function, we first check if $\textit{root}$ is null. If it is, we return $\text{false}$. Otherwise, we check if $\textit{root}$ and $\textit{subRoot}$ are identical. If they are, we return $\text{true}$. Otherwise, we recursively check if the left or right subtree of $\textit{root}$ contains $\textit{subRoot}$.

The time complexity is $O(n \times m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the number of nodes in the trees $root$ and $subRoot$, respectively.

Python3JavaC++GoTypeScriptRustJavaScript

# 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:
        def same(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
            if p is None or q is None:
                return p is q
            return p.val == q.val and same(p.left, q.left) and same(p.right, q.right)

        if root is None:
            return False
        return (
            same(root, subRoot)
            or self.isSubtree(root.left, subRoot)
            or self.isSubtree(root.right, subRoot)
        )

/**
 * 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;
        }
        return same(root, subRoot) || isSubtree(root.left, subRoot)
            || isSubtree(root.right, subRoot);
    }

    private boolean same(TreeNode p, TreeNode q) {
        if (p == null || q == null) {
            return p == q;
        }
        return p.val == q.val && same(p.left, q.left) && same(p.right, q.right);
    }
}

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

    bool same(TreeNode* p, TreeNode* q) {
        if (!p || !q) {
            return p == q;
        }
        return p->val == q->val && same(p->left, q->left) && same(p->right, q->right);
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSubtree(root *TreeNode, subRoot *TreeNode) bool {
    var same func(p, q *TreeNode) bool
    same = func(p, q *TreeNode) bool {
        if p == nil || q == nil {
            return p == q
        }
        return p.Val == q.Val && same(p.Left, q.Left) && same(p.Right, q.Right)
    }
    if root == nil {
        return false
    }
    return same(root, subRoot) || isSubtree(root.Left, subRoot) || isSubtree(root.Right, subRoot)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */
function isSubtree(root: TreeNode | null, subRoot: TreeNode | null): boolean {
    const same = (p: TreeNode | null, q: TreeNode | null): boolean => {
        if (!p || !q) {
            return p === q;
        }
        return p.val === q.val && same(p.left, q.left) && same(p.right, q.right);
    };
    if (!root) {
        return false;
    }
    return same(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn is_subtree(
        root: Option<Rc<RefCell<TreeNode>>>,
        sub_root: Option<Rc<RefCell<TreeNode>>>,
    ) -> bool {
        if root.is_none() {
            return false;
        }
        Self::same(&root, &sub_root)
            || Self::is_subtree(
                root.as_ref().unwrap().borrow().left.clone(),
                sub_root.clone(),
            )
            || Self::is_subtree(
                root.as_ref().unwrap().borrow().right.clone(),
                sub_root.clone(),
            )
    }

    fn same(p: &Option<Rc<RefCell<TreeNode>>>, q: &Option<Rc<RefCell<TreeNode>>>) -> bool {
        match (p, q) {
            (None, None) => true,
            (Some(p), Some(q)) => {
                let p = p.borrow();
                let q = q.borrow();
                p.val == q.val && Self::same(&p.left, &q.left) && Self::same(&p.right, &q.right)
            }
            _ => false,
        }
    }
}

/**
 * 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) {
    const same = (p, q) => {
        if (!p || !q) {
            return p === q;
        }
        return p.val === q.val && same(p.left, q.left) && same(p.right, q.right);
    };
    if (!root) {
        return false;
    }
    return same(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
};

572. Subtree of Another Tree

Description

Solutions

Solution 1: DFS

Comments