题目:
递归遍历,先序遍历,如果两个树相同,那么它们根节点一定相同并且它们的左子树相同,右子树也相同。
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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 isSameTree(TreeNode *p, TreeNode *q) { if (p == nullptr && q == nullptr) { return true; } if (p != nullptr && q != nullptr) { if (p->val != q->val) { return false; } else { return isSameTree(p->left, q->left) && isSameTree(p->right, q->right); } } return false; } };
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rust 可以一句话结束,因为 TreeNode 实现了 PartialEq, Eq, 🤣🤣🤣
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| #[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; struct Solution; impl Solution { pub fn is_same_tree( p: Option<Rc<RefCell<TreeNode>>>, q: Option<Rc<RefCell<TreeNode>>>, ) -> bool { p == q } } fn main() {}
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