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

863. All Nodes Distance K in Binary Tree
题目地址
​https://leetcode.com/problems/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);
​
        Queue<TreeNode> queue = new LinkedList();
        queue.add(target);
​
        Set<TreeNode> seen = new HashSet();
        seen.add(target);
      seen.add(null);
​
        int dist = 0;
        while (!queue.isEmpty()) {
            for (int size = queue.size(); size > 0; size--) {
                if (dist == K) {
                    for (TreeNode n: queue) {
                        ans.add(n.val);
                    }
          return ans;
                } else {
                    TreeNode node = queue.poll();
                    // left
                    if (!seen.contains(node.left)) {
                        seen.add(node.left);
                        queue.offer(node.left);
                    }
                    // right
                    if (!seen.contains(node.right)) {
                        seen.add(node.right);
                        queue.offer(node.right);
                    }
                    // up
                    TreeNode par = parent.get(node);
                    if (!seen.contains(par)) {
                        seen.add(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) {
        ans = new LinkedList();
        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) {
            subtree_add(node, 0);
            return 1;
        } else {
            int L = dfs(node.left);
            int R = dfs(node.right);
            if (L != -1) {
                if (L == K)     ans.add(node.val);
                subtree_add(node.right, L + 1);
                return L + 1;
            } else if (R != -1) {
                if (R == K)     ans.add(node.val);
                subtree_add(node.left, 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) {
            ans.add(node.val);
        } else {
            subtree_add(node.left, dist + 1);
            subtree_add(node.right, dist + 1);
        }
    }
}
854.K-Similar-Strings
864.Shortest-Path-to-Get-All-Keys
Last modified 2yr ago