Leetcode - Find Elements in a Contaminated Binary Tree

https://leetcode.com/problems/find-elements-in-a-contaminated-binary-tree/

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.

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
["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 

Contaminated only means it has a default value. No need to consider what kind of shit that it stores.

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


func Constructor(root *TreeNode) FindElements {
    hash := map[int]bool{}
    dfs(root, 0, &hash)
    return FindElements{
        root: root,
        hash: hash,
    }
}

func dfs(root *TreeNode, val int, hash *map[int]bool) {
    if root == nil {
        return
    }
    (*hash)[val] = true
    root.Val = val
    dfs(root.Left, 2*val+1, hash)
    dfs(root.Right, 2*val+2, hash)
}


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


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

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