Question

1261. 在受污染的二叉树中查找元素

English Version

题目描述

给出一个满足下述规则的二叉树：

1. root.val == 0
2. 如果 treeNode.val == x 且 treeNode.left != null，那么 treeNode.left.val == 2 * x + 1
3. 如果 treeNode.val == x 且 treeNode.right != null，那么 treeNode.right.val == 2 * x + 2

现在这个二叉树受到「污染」，所有的 treeNode.val 都变成了 -1。

请你先还原二叉树，然后实现 FindElements 类：

• FindElements(TreeNode* root) 用受污染的二叉树初始化对象，你需要先把它还原。
• bool find(int target) 判断目标值 target 是否存在于还原后的二叉树中并返回结果。

 

示例 1：

输入：
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
输出：
[null,false,true]
解释：
FindElements findElements = new FindElements([-1,null,-1]); 
findElements.find(1); // return False 
findElements.find(2); // return True 

示例 2：

输入：
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
输出：
[null,true,true,false]
解释：
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

示例 3：

输入：
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
输出：
[null,true,false,false,true]
解释：
FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]);
findElements.find(2); // return True
findElements.find(3); // return False
findElements.find(4); // return False
findElements.find(5); // return True

 

提示：

• TreeNode.val == -1
• 二叉树的高度不超过 20
• 节点的总数在 [1, 10^4] 之间
• 调用 find() 的总次数在 [1, 10^4] 之间
• 0 <= target <= 10^6

解法

方法一：DFS + 哈希表

我们先通过 DFS 遍历二叉树，将节点值恢复为原来的值，并将所有节点值存入哈希表中。然后在查找时，只需要判断哈希表中是否存在目标值即可。

时间复杂度方面，初始化时需要遍历二叉树，时间复杂度为 $O(n)$，查找时只需要判断哈希表中是否存在目标值，时间复杂度为 $O(1)$。空间复杂度 $O(n)$。其中 $n$ 为二叉树节点个数。

# 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 FindElements:

  def __init__(self, root: Optional[TreeNode]):
    def dfs(root: Optional[TreeNode]):
      self.s.add(root.val)
      if root.left:
        root.left.val = root.val * 2 + 1
        dfs(root.left)
      if root.right:
        root.right.val = root.val * 2 + 2
        dfs(root.right)

    root.val = 0
    self.s = set()
    dfs(root)

  def find(self, target: int) -> bool:
    return target in self.s


# Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)

/**
 * 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 FindElements {
  private Set<Integer> s = new HashSet<>();

  public FindElements(TreeNode root) {
    root.val = 0;
    dfs(root);
  }

  public boolean find(int target) {
    return s.contains(target);
  }

  private void dfs(TreeNode root) {
    s.add(root.val);
    if (root.left != null) {
      root.left.val = root.val * 2 + 1;
      dfs(root.left);
    }
    if (root.right != null) {
      root.right.val = root.val * 2 + 2;
      dfs(root.right);
    }
  }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * FindElements obj = new FindElements(root);
 * boolean param_1 = obj.find(target);
 */

/**
 * 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 FindElements {
public:
  FindElements(TreeNode* root) {
    root->val = 0;
    dfs(root);
  }

  bool find(int target) {
    return s.contains(target);
  }

private:
  unordered_set<int> s;

  void dfs(TreeNode* root) {
    s.insert(root->val);
    if (root->left) {
      root->left->val = root->val * 2 + 1;
      dfs(root->left);
    }
    if (root->right) {
      root->right->val = root->val * 2 + 2;
      dfs(root->right);
    }
  };
};

/**
 * Your FindElements object will be instantiated and called as such:
 * FindElements* obj = new FindElements(root);
 * bool param_1 = obj->find(target);
 */

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
type FindElements struct {
  s map[int]bool
}

func Constructor(root *TreeNode) FindElements {
  root.Val = 0
  s := map[int]bool{}
  var dfs func(*TreeNode)
  dfs = func(root *TreeNode) {
    s[root.Val] = true
    if root.Left != nil {
      root.Left.Val = root.Val*2 + 1
      dfs(root.Left)
    }
    if root.Right != nil {
      root.Right.Val = root.Val*2 + 2
      dfs(root.Right)
    }
  }
  dfs(root)
  return FindElements{s}
}

func (this *FindElements) Find(target int) bool {
  return this.s[target]
}

/**
 * Your FindElements object will be instantiated and called as such:
 * obj := Constructor(root);
 * param_1 := obj.Find(target);
 */

/**
 * 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)
 *   }
 * }
 */

class FindElements {
  private s: Set<number> = new Set<number>();

  constructor(root: TreeNode | null) {
    root.val = 0;
    const dfs = (root: TreeNode) => {
      this.s.add(root.val);
      if (root.left) {
        root.left.val = root.val * 2 + 1;
        dfs(root.left);
      }
      if (root.right) {
        root.right.val = root.val * 2 + 2;
        dfs(root.right);
      }
    };
    dfs(root);
  }

  find(target: number): boolean {
    return this.s.has(target);
  }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * var obj = new FindElements(root)
 * var param_1 = obj.find(target)
 */