LeetCode 863. All Nodes Distance K in Binary Tree

Tree

We are given a binary tree (with root node root), a target node, and an integer value K.

Return a list of the values of all nodes that have a distance K from the target node. The answer can be returned in any order.

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2

Output: [7,4,1]

Explanation: 
The nodes that are a distance 2 from the target node (with value 5)
have values 7, 4, and 1.



Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.

Note:

1. The given tree is non-empty.

2. Each node in the tree has unique values 0 <= node.val <= 500.

3. The target node is a node in the tree.

4. 0 <= K <= 1000.

Solution

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中文版解答Bilibili Link

// Solution 1: O(n) Time, O(n) Space
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> ans;   
    unordered_map<TreeNode*, TreeNode*> parent;  // son=>parent  
    unordered_set<TreeNode*> visit;    //record visied node
    
    void findParent(TreeNode* node) {
        if (node == nullptr) return;
        if (node->left != nullptr) {
            parent[node->left] = node;
            findParent(node->left);
        }
        if (node->right != nullptr){
            parent[node->right] = node;
            findParent(node->right);
        }
    }
    
    void dfs(TreeNode* node, int K){
        if (node == nullptr || visit.count(node) > 0) {
            return;
        }
        visit.insert(node);
        if (K == 0) {
            ans.push_back(node->val);
            return;
        }
        dfs(node->left, K-1);
        dfs(node->right, K-1);
        dfs(parent[node], K-1);
    }
    
    vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {
        findParent(root);
        dfs(target, K);
        return ans;
    }
};

// Solution 2: O(n) Time, O(lgn) Space
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    bool FindPath(TreeNode* root, TreeNode* target, vector<bool>& paths) {
        if (root == nullptr) {
            return false;
        }
        if (target == root) {
            return true;
        }
        
        bool l = FindPath(root->left, target, paths);
        if (l) {
            paths.push_back(false);
            return true;
        }
        bool r = FindPath(root->right, target, paths);
        if (r) {
            paths.push_back(true);
            return true;
        }
        
        return false;
    }
    
    void FindAnswer(TreeNode* root, vector<int>& ans, int dis, int K) {
        if (root == nullptr || dis > K) {
            return;
        }
        if (dis == K) {
            ans.push_back(root->val);
            return;
        }
        FindAnswer(root->left, ans, dis + 1, K);
        FindAnswer(root->right, ans, dis + 1, K);
    }
    
public:
    vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {
        vector<bool> paths;
        FindPath(root, target, paths);
        reverse(paths.begin(), paths.end());
        
        vector<int> ans;
        for (int i = 0; i < paths.size(); i++) {
            if (paths.size() - i == K) {
                ans.push_back(root->val);
            }
            bool path = paths[i];
            if (path == false) {
                FindAnswer(root->right, ans, paths.size() + 1 - i, K);
                root = root->left;
            } else {
                FindAnswer(root->left, ans, paths.size() + 1 - i, K);
                root = root->right;
            }
        }
        FindAnswer(root, ans, 0, K);
        return ans;
    }
};