# 30 Day Challenge Day 21 | Leetcode 863. All Nodes Distance K in Binary Tree

## 题解

Medium

BFS

Tree 是一种特殊的 Graph，节点之间只有一个方向。而这里有从 Target 节点向各个方向，包括父节点方向，遍历的需求。所以利用 Hashmap 先转化成一个无向图。

/**
* 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> distanceK(TreeNode* root, TreeNode* target, int K) {
vector<int> ret;

// convert to bi-directional graph
unordered_map<TreeNode*, vector<TreeNode*>> graph;
unordered_set<TreeNode*> visited; // avoid cycle traversal

queue<TreeNode*> q;
q.push(root);

while(!q.empty()) {
auto t = q.front();
q.pop();

if(t->left) {
graph[t].push_back(t->left);
graph[t->left].push_back(t);
q.push(t->left);
}
if(t->right) {
graph[t].push_back(t->right);
graph[t->right].push_back(t);
q.push(t->right);
}
}

// start from target
q.push(target);

while(!q.empty() && K >= 0) {
int sz = q.size();
for(int i = 0; i < sz; i++) {
auto t = q.front();
q.pop();

visited.insert(t);

if(K == 0) {
ret.push_back(t->val);
}

for(auto node : graph[t]) {
if(!visited.count(node)) q.push(node);
}
}
K--;
}

return ret;
}
};

posted @ 2020-10-09 07:30  CasperWin  阅读(129)  评论(0编辑  收藏  举报