[数据结构与算法]LeetCode每日一题(Binary Tree Pruning)

Given the root of a binary tree, return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

A subtree of a node node is node plus every node that is a descendant of node.

Example 1:

Input: root = [1,null,0,0,1]
Output: [1,null,0,null,1]
Explanation:
Only the red nodes satisfy the property “every subtree not containing a 1”.
The diagram on the right represents the answer.
Example 2:

Input: root = [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]
Example 3:

Input: root = [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]

Constraints:

The number of nodes in the tree is in the range [1, 200].
Node.val is either 0 or 1.

就两种情况

1. 节点值是1的话，将左右节点进行筛选，然后返回当前节点
2. 节点值是0的话，如果左右节点都是0，则返回空值，否则与节点值是1的情况相同

代码(Rust):

impl Solution {
    fn rc(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        if let Some(node) = root {
            if node.borrow().val == 1 {
                let mut left = node.borrow_mut().left.take();
                let mut right = node.borrow_mut().right.take();
                left = Solution::rc(left);
                right = Solution::rc(right);
                node.borrow_mut().left = left;
                node.borrow_mut().right = right;
                return Some(node);
            } else {
                let left = Solution::rc(node.borrow_mut().left.take());
                let right = Solution::rc(node.borrow_mut().right.take());
                if left.is_none() && right.is_none() {
                    return None;
                }
                node.borrow_mut().left = left;
                node.borrow_mut().right = right;
                return Some(node);
            }
        } else {
            return None;
        }
    }

    pub fn prune_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        Solution::rc(root)
    }
}