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
# 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: TreeNode, subRoot: TreeNode) -> bool:
def dfs(root1, root2):
if root1 is None and root2 is None:
return True
if root1 is None or root2 is None:
return False
return (
root1.val == root2.val
and dfs(root1.left, root2.left)
and dfs(root1.right, root2.right)
)
if root is None:
return False
return (
dfs(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 dfs(root, subRoot) || isSubtree(root.left, subRoot)
|| isSubtree(root.right, subRoot);
}
private boolean dfs(TreeNode root1, TreeNode root2) {
if (root1 == null && root2 == null) {
return true;
}
if (root1 == null || root2 == null) {
return false;
}
return root1.val == root2.val && dfs(root1.left, root2.left)
&& dfs(root1.right, root2.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 0;
return dfs(root, subRoot) || isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot);
}
bool dfs(TreeNode* root1, TreeNode* root2) {
if (!root1 && !root2) return 1;
if (!root1 || !root2) return 0;
return root1->val == root2->val && dfs(root1->left, root2->left) && dfs(root1->right, root2->right);
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isSubtree(root *TreeNode, subRoot *TreeNode) bool {
if root == nil {
return false
}
var dfs func(root1, root2 *TreeNode) bool
dfs = func(root1, root2 *TreeNode) bool {
if root1 == nil && root2 == nil {
return true
}
if root1 == nil || root2 == nil {
return false
}
return root1.Val == root2.Val && dfs(root1.Left, root2.Left) && dfs(root1.Right, root2.Right)
}
return dfs(root, subRoot) || isSubtree(root.Left, subRoot) || isSubtree(root.Right, subRoot)
}/**
* 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;
let dfs = function (root1, root2) {
if (!root1 && !root2) {
return true;
}
if (!root1 || !root2) {
return false;
}
return (
root1.val == root2.val && dfs(root1.left, root2.left) && dfs(root1.right, root2.right)
);
};
return dfs(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)
* }
* }
*/
const dfs = (root: TreeNode | null, subRoot: TreeNode | null) => {
if (root == null && subRoot == null) {
return true;
}
if (root == null || subRoot == null || root.val !== subRoot.val) {
return false;
}
return dfs(root.left, subRoot.left) && dfs(root.right, subRoot.right);
};
function isSubtree(root: TreeNode | null, subRoot: TreeNode | null): boolean {
if (root == null) {
return false;
}
return dfs(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::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, sub_root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if root.is_none() && sub_root.is_none() {
return true;
}
if root.is_none() || sub_root.is_none() {
return false;
}
let root = root.as_ref().unwrap().borrow();
let sub_root = sub_root.as_ref().unwrap().borrow();
root.val == sub_root.val
&& Self::dfs(&root.left, &sub_root.left)
&& Self::dfs(&root.right, &sub_root.right)
}
fn help(
root: &Option<Rc<RefCell<TreeNode>>>,
sub_root: &Option<Rc<RefCell<TreeNode>>>,
) -> bool {
if root.is_none() {
return false;
}
Self::dfs(root, sub_root)
|| Self::help(&root.as_ref().unwrap().borrow().left, sub_root)
|| Self::help(&root.as_ref().unwrap().borrow().right, sub_root)
}
pub fn is_subtree(
root: Option<Rc<RefCell<TreeNode>>>,
sub_root: Option<Rc<RefCell<TreeNode>>>,
) -> bool {
Self::help(&root, &sub_root)
}
}