1261. Find Elements in a Contaminated Binary Tree

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.

Python3JavaC++GoTypeScript

# 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)
 */

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