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1261. 在受污染的二叉树中查找元素

English Version

题目描述

给出一个满足下述规则的二叉树：

  1. root.val == 0
  2. 如果 treeNode.val == x 且 treeNode.left != null，那么 treeNode.left.val == 2 * x + 1
  3. 如果 treeNode.val == x 且 treeNode.right != null，那么 treeNode.right.val == 2 * x + 2

现在这个二叉树受到「污染」，所有的 treeNode.val 都变成了 -1。

请你先还原二叉树，然后实现 FindElements 类：

  • FindElements(TreeNode* root) 用受污染的二叉树初始化对象，你需要先把它还原。
  • bool find(int target) 判断目标值 target 是否存在于还原后的二叉树中并返回结果。

 

示例 1：

输入：
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
输出：
[null,false,true]
解释：
FindElements findElements = new FindElements([-1,null,-1]); 
findElements.find(1); // return False 
findElements.find(2); // return True

示例 2：

输入：
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
输出：
[null,true,true,false]
解释：
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

示例 3：

输入：
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
输出：
[null,true,false,false,true]
解释：
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

 

提示：

  • TreeNode.val == -1
  • 二叉树的高度不超过 20
  • 节点的总数在 [1, 10^4] 之间
  • 调用 find() 的总次数在 [1, 10^4] 之间
  • 0 <= target <= 10^6

解法

方法一：DFS + 哈希表

我们先通过 DFS 遍历二叉树，将节点值恢复为原来的值，然后再通过哈希表存储所有节点值，这样在查找时就可以直接判断目标值是否存在于哈希表中。

时间复杂度方面，初始化时需要遍历二叉树，时间复杂度为 $O(n)$，查找时只需要判断哈希表中是否存在目标值，时间复杂度为 $O(1)$。空间复杂度 $O(n)$。其中 $n$ 为二叉树节点个数。

Python3

# 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):
            self.vis.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.vis = set()
        dfs(root)

    def find(self, target: int) -> bool:
        return target in self.vis


# Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)

Java

/**
 * 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> vis = new HashSet<>();

    public FindElements(TreeNode root) {
        root.val = 0;
        dfs(root);
    }

    private void dfs(TreeNode root) {
        vis.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);
        }
    }

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

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

C++

/**
 * 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;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            vis.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);
            }
        };
        dfs(root);
    }

    bool find(int target) {
        return vis.count(target);
    }

private:
    unordered_set<int> vis;
};

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

Go

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

func Constructor(root *TreeNode) FindElements {
	root.Val = 0
	vis := map[int]bool{}
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		vis[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{vis}
}

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

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

...