Question

814. 二叉树剪枝

English Version

题目描述

给你二叉树的根结点 root ，此外树的每个结点的值要么是 0 ，要么是 1 。

返回移除了所有不包含 1 的子树的原二叉树。

节点 node 的子树为 node 本身加上所有 node 的后代。

 

示例 1：

输入：root = [1,null,0,0,1]
输出：[1,null,0,null,1]
解释：
只有红色节点满足条件“所有不包含 1 的子树”。 右图为返回的答案。

示例 2：

输入：root = [1,0,1,0,0,0,1]
输出：[1,null,1,null,1]

示例 3：

输入：root = [1,1,0,1,1,0,1,0]
输出：[1,1,0,1,1,null,1]

 

提示：

• 树中节点的数目在范围 [1, 200] 内
• Node.val 为 0 或 1

解法

方法一：DFS

观察叶节点，当叶节点 val 为 0 时，便将该节点抹去。回溯，查看其父节点是否成为了新的叶节点，依照此规则自底向上。

# 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 pruneTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
    if root is None:
      return None
    root.left = self.pruneTree(root.left)
    root.right = self.pruneTree(root.right)
    if root.val == 0 and root.left is None and root.right is None:
      return None
    return root

/**
 * 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 TreeNode pruneTree(TreeNode root) {
    if (root == null) {
      return null;
    }
    root.left = pruneTree(root.left);
    root.right = pruneTree(root.right);
    if (root.val == 0 && root.left == null && root.right == null) {
      return null;
    }
    return root;
  }
}

/**
 * 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:
  TreeNode* pruneTree(TreeNode* root) {
    if (!root) return nullptr;
    root->left = pruneTree(root->left);
    root->right = pruneTree(root->right);
    if (!root->val && !root->left && !root->right) return nullptr;
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func pruneTree(root *TreeNode) *TreeNode {
  if root == nil {
    return nil
  }
  root.Left = pruneTree(root.Left)
  root.Right = pruneTree(root.Right)
  if root.Val == 0 && root.Left == nil && root.Right == nil {
    return nil
  }
  return root
}

/**
 * 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 pruneTree(root: TreeNode | null): TreeNode | null {
  if (root == null) {
    return root;
  }
  root.left = pruneTree(root.left);
  root.right = pruneTree(root.right);
  if (root.val == 0 && root.left == null && root.right == null) {
    return null;
  }
  return root;
}

// 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 {
  pub fn prune_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
    if root.is_none() {
      return None;
    }

    let root = root.unwrap();
    let left = Self::prune_tree(root.borrow_mut().left.take());
    let right = Self::prune_tree(root.borrow_mut().right.take());
    if root.borrow().val == 0 && left.is_none() && right.is_none() {
      return None;
    }

    root.borrow_mut().left = left;
    root.borrow_mut().right = right;
    Some(root)
  }
}

/**
 * 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
 * @return {TreeNode}
 */
var pruneTree = function (root) {
  if (!root) return null;
  root.left = pruneTree(root.left);
  root.right = pruneTree(root.right);
  if (root.val == 0 && !root.left && !root.right) {
    return null;
  }
  return root;
};