Find elements in a contaminated binary tree [DFS]

Time: O(N); Space: O(H); medium

Given a binary tree with the following rules:

  • root.val == 0

  • If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1

  • 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.

You need to first recover the binary tree and then implement the FindElements class:

  • FindElements(TreeNode* root) Initializes the object with a contamined binary tree, you need to recover it first.

  • bool find(int target) Return if the target value exists in the recovered binary tree.

Example 1:

Input/Output:

  • root = {TreeNode} [-1,null,-1]

  • f = FindElements(root)

  • f.find(1) -> False

  • f.find(2) -> True

Example 2:

Input/Output:

  • root = {TreeNode} [-1,-1,-1,-1,-1]

  • f = FindElements(root)

  • f.find(1) -> True

  • f.find(3) -> True

  • f.find(5) -> False

Example 3:

Input/Output:

  • root = {TreeNode} [-1,null,-1,-1,null,-1]

  • f = FindElements(root)

  • f.find(2) -> True

  • f.find(3) -> False

  • f.find(4) -> False

  • f.find(5) -> 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, 10^4]

  • Total calls of find() is between [1, 10^4]

  • 0 <= target <= 10^6

Hints:

  1. Use DFS to traverse the binary tree and recover it.

  2. Use a hashset to store TreeNode.val for finding.

[25]:
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right