跳转至

863. 二叉树中所有距离为 K 的结点

题目描述

给定一个二叉树（具有根结点 root）， 一个目标结点 target ，和一个整数值 k ，返回到目标结点 target 距离为 k 的所有结点的值的数组。

答案可以以 任何顺序 返回。

 

示例 1：

输入：root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
输出：[7,4,1]
解释：所求结点为与目标结点（值为 5）距离为 2 的结点，值分别为 7，4，以及 1

示例 2:

输入: root = [1], target = 1, k = 3
输出: []

 

提示:

  • 节点数在 [1, 500] 范围内
  • 0 <= Node.val <= 500
  • Node.val 中所有值 不同
  • 目标结点 target 是树上的结点。
  • 0 <= k <= 1000

 

解法

方法一：DFS + 哈希表

我们先用 DFS 遍历整棵树，记录每个结点的父结点，然后从目标结点开始，向上、向下分别搜索距离为 $k$ 的结点，添加到答案数组中。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的结点数。

 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
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
        def parents(root, prev):
            nonlocal p
            if root is None:
                return
            p[root] = prev
            parents(root.left, root)
            parents(root.right, root)

        def dfs(root, k):
            nonlocal ans, vis
            if root is None or root.val in vis:
                return
            vis.add(root.val)
            if k == 0:
                ans.append(root.val)
                return
            dfs(root.left, k - 1)
            dfs(root.right, k - 1)
            dfs(p[root], k - 1)

        p = {}
        parents(root, None)
        ans = []
        vis = set()
        dfs(target, k)
        return ans

 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
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private Map<TreeNode, TreeNode> p;
    private Set<Integer> vis;
    private List<Integer> ans;

    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        p = new HashMap<>();
        vis = new HashSet<>();
        ans = new ArrayList<>();
        parents(root, null);
        dfs(target, k);
        return ans;
    }

    private void parents(TreeNode root, TreeNode prev) {
        if (root == null) {
            return;
        }
        p.put(root, prev);
        parents(root.left, root);
        parents(root.right, root);
    }

    private void dfs(TreeNode root, int k) {
        if (root == null || vis.contains(root.val)) {
            return;
        }
        vis.add(root.val);
        if (k == 0) {
            ans.add(root.val);
            return;
        }
        dfs(root.left, k - 1);
        dfs(root.right, k - 1);
        dfs(p.get(root), k - 1);
    }
}

 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
/**
 * 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:
    unordered_map<TreeNode*, TreeNode*> p;
    unordered_set<int> vis;
    vector<int> ans;

    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        parents(root, nullptr);
        dfs(target, k);
        return ans;
    }

    void parents(TreeNode* root, TreeNode* prev) {
        if (!root) return;
        p[root] = prev;
        parents(root->left, root);
        parents(root->right, root);
    }

    void dfs(TreeNode* root, int k) {
        if (!root || vis.count(root->val)) return;
        vis.insert(root->val);
        if (k == 0) {
            ans.push_back(root->val);
            return;
        }
        dfs(root->left, k - 1);
        dfs(root->right, k - 1);
        dfs(p[root], k - 1);
    }
};

 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
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func distanceK(root *TreeNode, target *TreeNode, k int) []int {
    p := make(map[*TreeNode]*TreeNode)
    vis := make(map[int]bool)
    var ans []int
    var parents func(root, prev *TreeNode)
    parents = func(root, prev *TreeNode) {
        if root == nil {
            return
        }
        p[root] = prev
        parents(root.Left, root)
        parents(root.Right, root)
    }
    parents(root, nil)
    var dfs func(root *TreeNode, k int)
    dfs = func(root *TreeNode, k int) {
        if root == nil || vis[root.Val] {
            return
        }
        vis[root.Val] = true
        if k == 0 {
            ans = append(ans, root.Val)
            return
        }
        dfs(root.Left, k-1)
        dfs(root.Right, k-1)
        dfs(p[root], k-1)
    }
    dfs(target, k)
    return ans
}

方法二

 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
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
        def dfs1(root, fa):
            if root is None:
                return
            p[root] = fa
            dfs1(root.left, root)
            dfs1(root.right, root)

        def dfs2(root, fa, k):
            if root is None:
                return
            if k == 0:
                ans.append(root.val)
                return
            for nxt in (root.left, root.right, p[root]):
                if nxt != fa:
                    dfs2(nxt, root, k - 1)

        p = {}
        dfs1(root, None)
        ans = []
        dfs2(target, None, k)
        return ans

评论