LC.P863[二叉树中所有距离为 K 的结点]

思路

树是一类特殊的图，对二叉树中相邻的节点建图，边的权重为1。接着从目标节点出发遍历图，找到距离为k的节点。

方法一：建图+BFS

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int N = 510, M = 4 * N, idx = 0;
    int[] he = new int[N], e = new int[M], ne = new int[M];
    boolean[] visited = new boolean[N];

    private void add(int a, int b) {
        e[idx] = b;
        ne[idx] = he[a];
        he[a] = idx++;
    }

    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        Arrays.fill(he, -1);
        dfs(root);
        List<Integer> ans = new ArrayList<>();
        Deque<Integer> queue = new ArrayDeque<>();
        queue.offer(target.val);
        visited[target.val] = true;
        while (!queue.isEmpty() && k >= 0) {
            int size = queue.size();
            while (size-- > 0) {
                int cur = queue.poll();
                if (k == 0) {
                    ans.add(cur);
                    continue;
                }
                for (int i = he[cur]; i != -1; i = ne[i]) {
                    int j = e[i];
                    if (!visited[j]) {
                        queue.offer(j);
                        visited[j] = true;
                    }
                }
            }
            --k;
        }
        return ans;
    }

    private void dfs(TreeNode node) {
        if (node == null) return;
        if (node.left != null) {
            add(node.val, node.left.val);
            add(node.left.val, node.val);
            dfs(node.left);
        }
        if (node.right != null) {
            add(node.val, node.right.val);
            add(node.right.val, node.val);
            dfs(node.right);
        }
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：建图+迭代

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int N = 510, M = 4 * N, idx = 0;
    int[] he = new int[N], e = new int[M], ne = new int[M];
    boolean[] visited = new boolean[N];
    List<Integer> ans;

    private void add(int a, int b) {
        e[idx] = b;
        ne[idx] = he[a];
        he[a] = idx++;
    }

    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        ans = new ArrayList<>();
        Arrays.fill(he, -1);
        dfs(root);
        visited[target.val] = true;
        find(target.val, k, 0);
        return ans;
    }

    private void find(int node, int k, int cur) {
        if (cur == k) {
            ans.add(node);
            return;
        }
        for (int i = he[node]; i != -1; i = ne[i]) {
            int j = e[i];
            if (!visited[j]) {
                visited[j] = true;
                find(j, k, cur + 1);
            }
        }
    }

    private void dfs(TreeNode node) {
        if (node == null) return;
        if (node.left != null) {
            add(node.val, node.left.val);
            add(node.left.val, node.val);
            dfs(node.left);
        }
        if (node.right != null) {
            add(node.val, node.right.val);
            add(node.right.val, node.val);
            dfs(node.right);
        }
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法三：建图(map)+迭代

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    Map<Integer, List<Integer>> adj = new HashMap<>();
    boolean[] visited = new boolean[510];
    List<Integer> ans = new ArrayList<>();

    private void add(int a, int b) {
        adj.computeIfAbsent(a, key -> new ArrayList<>()).add(b);
    }

    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        dfs(root);
        visited[target.val] = true;
        find(target.val, k, 0);
        return ans;
    }

    private void find(int node, int k, int cur) {
        if (cur == k) {
            ans.add(node);
            return;
        }
        List<Integer> list = adj.get(node);
        if (list == null) return;
        for (Integer i : list) {
            if (!visited[i]) {
                visited[i] = true;
                find(i, k, cur + 1);
            }
        }
    }

    private void dfs(TreeNode node) {
        if (node == null) return;
        if (node.left != null) {
            add(node.val, node.left.val);
            add(node.left.val, node.val);
            dfs(node.left);
        }
        if (node.right != null) {
            add(node.val, node.right.val);
            add(node.right.val, node.val);
            dfs(node.right);
        }
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$