1261. 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 

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, 10^4]
• Total calls of find() is between [1, 10^4]
• 0 <= target <= 10^6

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class FindElements {
    TreeNode t3;
    public FindElements(TreeNode root) {
        TreeNode t = new TreeNode(0);
        t3 = t;
        F(root, t);
        
    }
    
    public void F(TreeNode root, TreeNode t){
        if(root == null) return;
        if(root.left != null){
            TreeNode t1 = new TreeNode(t.val * 2 + 1);
            t.left = t1;
            F(root.left, t.left);
        }
        if(root.right != null){
            TreeNode t2 = new TreeNode(t.val * 2 + 2);
            t.right = t2;
            F(root.right, t.right);
        }
    }
    public boolean find(int target) {
       
       return find1(t3, target);
       
        //return false;
    }
    public boolean find1(TreeNode x, int target){
        if(x == null) return false;
        if(x.val == target) return true;
        return find1(x.left, target) || find1(x.right, target);
    }
}

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

这题巩固了几个知识点：

1. 克隆tree，写两个方法，一个初始化根的value，另一个void方法不停的call复制树和被复制树.也是如果null就return，不然就判断left和right是否为空，不空就新建node，连接，然后recursively call

2.查找tree的元素，如果null就false，如果相等就true，否则查找左右子树。

class FindElements {

        Set<Integer> set;

        /**
         * Build the tree using dfs, put all values to the set. Then for find use set for lookups
         */
        public FindElements(TreeNode root) {
            set = new HashSet();
            dfs(root, 0);
        }

        void dfs(TreeNode n, int val) {
            if (n == null)
                return;

            set.add(val);
            dfs(n.left, 2*val + 1);
            dfs(n.right, 2*val + 2);
        }

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

大佬的方法，用set存value，屌

class FindElements {
    TreeNode t3;
    Set<Integer> set;
   
    public FindElements(TreeNode root) {
        TreeNode t = new TreeNode(0);
        set = new HashSet();
        set.add(0);
        t3 = t;
        F(root, t);
    }
    
    public void F(TreeNode root, TreeNode t){
        if(root == null) return;
        if(root.left != null){
            TreeNode t1 = new TreeNode(t.val * 2 + 1);
            set.add(t.val * 2 + 1);
            t.left = t1;
            F(root.left, t.left);
        }
        if(root.right != null){
            TreeNode t2 = new TreeNode(t.val * 2 + 2);
            set.add(t.val * 2 + 2);
            t.right = t2;
            F(root.right, t.right);
        }
    }
    public boolean find(int target) {
       return set.contains(target);
    }

}

再修改一下我的。

1261. Find Elements in a Contaminated Binary Tree

原文：https://www.cnblogs.com/wentiliangkaihua/p/11886734.html