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# 863.All-Nodes-Distance-K-in-Binary-Tree

## 题目描述

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:
The given tree is non-empty.
Each node in the tree has unique values 0 <= node.val <= 500.
The target node is a node in the tree.
0 <= K <= 1000.

## 代码

### Approach #1 Parent Map

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
Map<TreeNode, TreeNode> parent;
public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
List<Integer> ans = new ArrayList();
if (root == null) return ans;
parent = new HashMap();
preorder(root, null);
Set<TreeNode> seen = new HashSet();
int dist = 0;
while (!queue.isEmpty()) {
for (int size = queue.size(); size > 0; size--) {
if (dist == K) {
for (TreeNode n: queue) {
}
return ans;
} else {
TreeNode node = queue.poll();
// left
if (!seen.contains(node.left)) {
queue.offer(node.left);
}
// right
if (!seen.contains(node.right)) {
queue.offer(node.right);
}
// up
TreeNode par = parent.get(node);
if (!seen.contains(par)) {
queue.offer(par);
}
}
}
dist++;
}
return ans;
}
public void preorder(TreeNode node, TreeNode par) {
if (node != null) {
parent.put(node, par);
preorder(node.left, node);
preorder(node.right, node);
}
}
}

### Approach #2 Percolate Distance

Algorithm
Traverse every `node` with a depth first search `dfs`. We'll add all nodes `x` to the answer such that `node` is the node on the path from `x` to `target` that is closest to the `root`.
To help us, `dfs(node)` will return the distance from `node` to the `target`. Then, there are 4 cases:
• If `node == target`, then we should add nodes that are distance `K` in the subtree rooted at `target`.
• If `target` is in the left branch of `node`, say at distance `L+1`, then we should look for nodes that are distance `K - L - 1` in the right branch.
• If `target` is in the right branch of `node`, the algorithm proceeds similarly.
• If `target` isn't in either branch of `node`, then we stop.
Complexity Analysis
• Time Complexity: O_(_N), where N is the number of nodes in the given tree.
• Space Complexity: O_(_N).
class Solution {
List<Integer> ans;
TreeNode target;
int K;
public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
this.target = target;
this.K = K;
dfs(root);
return ans;
}
// Return vertex distance from node to target if exists, else -1
// Vertex distance: the number of vertices on the path from node to target
public int dfs(TreeNode node) {
if (node == null) {
return -1;
} else if (node == target) {
return 1;
} else {
int L = dfs(node.left);
int R = dfs(node.right);
if (L != -1) {
return L + 1;
} else if (R != -1) {
return R + 1;
} else {
return -1;
}
}
}
// Add all nodes 'K - dist' from the node to answer.
public void subtree_add(TreeNode node, int dist) {
if (node == null) return;
if (dist == K) {