题解|#判断t1树中是否有与t2树完全相同的子树#|Rust
判断t1树中是否有与t2树完全相同的子树
https://www.nowcoder.com/practice/4eaccec5ee8f4fe8a4309463b807a542
/**
* #[derive(PartialEq, Eq, Debug, Clone)]
* pub struct TreeNode {
* pub val: i32,
* pub left: Option<Box<TreeNode>>,
* pub right: Option<Box<TreeNode>>,
* }
*
* impl TreeNode {
* #[inline]
* fn new(val: i32) -> Self {
* TreeNode {
* val: val,
* left: None,
* right: None,
* }
* }
* }
*/
struct Solution{
}
impl Solution {
fn new() -> Self {
Solution{}
}
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root1 TreeNode类
* @param root2 TreeNode类
* @return bool布尔型
*/
pub fn isContains(&self, root1: Option<Box<TreeNode>>, root2: Option<Box<TreeNode>>) -> bool {
if root1.is_none() && root2.is_none() {
return true;
}
if root1.is_none() || root2.is_none() {
return false;
}
return Solution::dfs(self, Solution::find(self, root1, root2.as_ref().unwrap().val), root2);
}
fn find(&self, mut tree1: Option<Box<TreeNode>>, val: i32) -> Option<Box<TreeNode>> {
if tree1.is_none() {
return None
}
if tree1.as_ref().unwrap().val == val {
return tree1;
}
let mut ans = Solution::find(self, tree1.as_mut().unwrap().left.take(), val);
if ans.is_none() {
ans = Solution::find(self, tree1.as_mut().unwrap().right.take(), val);
}
return ans;
}
fn dfs(&self, mut tree1: Option<Box<TreeNode>>, mut tree2: Option<Box<TreeNode>>) -> bool {
if tree1.is_none() && tree2.is_none() {
return true;
}
if tree1.is_none() || tree2.is_none() {
return false;
}
if tree1.as_ref().unwrap().val != tree2.as_ref().unwrap().val {
return false;
}
return Solution::dfs(self, tree1.as_mut().unwrap().left.take(), tree2.as_mut().unwrap().left.take()) &&
Solution::dfs(self, tree1.as_mut().unwrap().right.take(), tree2.as_mut().unwrap().right.take());
}
}
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