Question

1261. Find Elements in a Contaminated Binary Tree

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Description

Given a binary tree with the following rules:

1. root.val == 0
2. If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
3. If treeNode.val == x and treeNode.right != null, then treeNode.right.val == 2 * x + 2

Now the binary tree is contaminated, which means all treeNode.val have been changed to -1.

Implement the FindElements class:

• FindElements(TreeNode* root) Initializes the object with a contaminated binary tree and recovers it.
• bool find(int target) Returns true if the target value exists in the recovered binary tree.

 

Example 1:

Input
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
Output
[null,false,true]
Explanation
FindElements findElements = new FindElements([-1,null,-1]); 
findElements.find(1); // return False 
findElements.find(2); // return True 

Example 2:

Input
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
Output
[null,true,true,false]
Explanation
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

Example 3:

Input
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
Output
[null,true,false,false,true]
Explanation
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

 

Constraints:

• TreeNode.val == -1
• The height of the binary tree is less than or equal to 20
• The total number of nodes is between [1, 104]
• Total calls of find() is between [1, 104]
• 0 <= target <= 106

Solutions

Solution 1: DFS + Hash Table

First, we traverse the binary tree using DFS, restore the node values to their original values, and store all node values in a hash table. Then, when searching, we only need to check if the target value exists in the hash table.

In terms of time complexity, it takes $O(n)$ time to traverse the binary tree during initialization, and $O(1)$ time to check if the target value exists in the hash table during search. The space complexity is $O(n)$, where $n$ is the number of nodes in the binary tree.

# 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)
 */