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简单
树
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二叉树
字符串匹配
哈希函数

572. 另一棵树的子树

English Version

题目描述

给你两棵二叉树 root 和 subRoot 。检验 root 中是否包含和 subRoot 具有相同结构和节点值的子树。如果存在，返回 true ；否则，返回 false 。

二叉树 tree 的一棵子树包括 tree 的某个节点和这个节点的所有后代节点。tree 也可以看做它自身的一棵子树。

 

示例 1：

输入：root = [3,4,5,1,2], subRoot = [4,1,2]
输出：true

示例 2：

输入：root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
输出：false

 

提示：

  • root 树上的节点数量范围是 [1, 2000]
  • subRoot 树上的节点数量范围是 [1, 1000]
  • -104 <= root.val <= 104
  • -104 <= subRoot.val <= 104

解法

方法一：DFS

我们定义一个辅助函数 $\textit{same}(p, q)$，用于判断以 $p$ 为根节点的树和以 $q$ 为根节点的树是否相等。如果两棵树的根节点的值相等，并且它们的左子树和右子树也分别相等，那么这两棵树是相等的。

在 $\textit{isSubtree}(\textit{root}, \textit{subRoot})$ 函数中，我们首先判断 $\textit{root}$ 是否为空，如果为空，则返回 $\text{false}$。否则，我们判断 $\textit{root}$ 和 $\textit{subRoot}$ 是否相等，如果相等，则返回 $\text{true}$。否则，我们递归地判断 $\textit{root}$ 的左子树和右子树是否包含 $\textit{subRoot}$。

时间复杂度 $O(n \times m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是树 $root$ 和树 $subRoot$ 的节点个数。

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:
        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)
        )

Java

/**
 * 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);
    }
}

C++

/**
 * 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);
    }
};

Go

/**
 * 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)
}

TypeScript

/**
 * 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);
}

Rust

// 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,
        }
    }
}

JavaScript

/**
 * 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);
};