跳转至

653. 两数之和 IV - 输入二叉搜索树

题目描述

给定一个二叉搜索树 root 和一个目标结果 k，如果二叉搜索树中存在两个元素且它们的和等于给定的目标结果，则返回 true。

 

示例 1：

输入: root = [5,3,6,2,4,null,7], k = 9
输出: true

示例 2：

输入: root = [5,3,6,2,4,null,7], k = 28
输出: false

 

提示:

  • 二叉树的节点个数的范围是  [1, 104].
  • -104 <= Node.val <= 104
  • 题目数据保证，输入的 root 是一棵 有效 的二叉搜索树
  • -105 <= k <= 105

解法

方法一：哈希表 + DFS

DFS 遍历二叉搜索树，对于每个节点，判断 k - node.val 是否在哈希表中，如果在，则返回 true，否则将 node.val 加入哈希表中。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉搜索树的节点个数。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findTarget(self, root: Optional[TreeNode], k: int) -> bool:
        def dfs(root):
            if root is None:
                return False
            if k - root.val in vis:
                return True
            vis.add(root.val)
            return dfs(root.left) or dfs(root.right)

        vis = set()
        return dfs(root)

 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
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Set<Integer> vis = new HashSet<>();
    private int k;

    public boolean findTarget(TreeNode root, int k) {
        this.k = k;
        return dfs(root);
    }

    private boolean dfs(TreeNode root) {
        if (root == null) {
            return false;
        }
        if (vis.contains(k - root.val)) {
            return true;
        }
        vis.add(root.val);
        return dfs(root.left) || dfs(root.right);
    }
}

 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
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool findTarget(TreeNode* root, int k) {
        unordered_set<int> vis;

        function<bool(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return false;
            }
            if (vis.count(k - root->val)) {
                return true;
            }
            vis.insert(root->val);
            return dfs(root->left) || dfs(root->right);
        };
        return dfs(root);
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findTarget(root *TreeNode, k int) bool {
    vis := map[int]bool{}
    var dfs func(*TreeNode) bool
    dfs = func(root *TreeNode) bool {
        if root == nil {
            return false
        }
        if vis[k-root.Val] {
            return true
        }
        vis[root.Val] = true
        return dfs(root.Left) || dfs(root.Right)
    }
    return dfs(root)
}

 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
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function findTarget(root: TreeNode | null, k: number): boolean {
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return false;
        }
        if (vis.has(k - root.val)) {
            return true;
        }
        vis.add(root.val);
        return dfs(root.left) || dfs(root.right);
    };
    const vis = new Set<number>();
    return dfs(root);
}

 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.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::{HashSet, VecDeque};
use std::rc::Rc;
impl Solution {
    pub fn find_target(root: Option<Rc<RefCell<TreeNode>>>, k: i32) -> bool {
        let mut set = HashSet::new();
        let mut q = VecDeque::new();
        q.push_back(root);
        while let Some(node) = q.pop_front() {
            if let Some(node) = node {
                let mut node = node.as_ref().borrow_mut();
                if set.contains(&node.val) {
                    return true;
                }
                set.insert(k - node.val);
                q.push_back(node.left.take());
                q.push_back(node.right.take());
            }
        }
        false
    }
}

方法二：哈希表 + BFS

与方法一类似，只是使用 BFS 遍历二叉搜索树。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉搜索树的节点个数。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findTarget(self, root: Optional[TreeNode], k: int) -> bool:
        q = deque([root])
        vis = set()
        while q:
            for _ in range(len(q)):
                node = q.popleft()
                if k - node.val in vis:
                    return True
                vis.add(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return False

 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
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean findTarget(TreeNode root, int k) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        Set<Integer> vis = new HashSet<>();
        while (!q.isEmpty()) {
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.poll();
                if (vis.contains(k - node.val)) {
                    return true;
                }
                vis.add(node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return false;
    }
}

 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
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool findTarget(TreeNode* root, int k) {
        queue<TreeNode*> q{{root}};
        unordered_set<int> vis;
        while (!q.empty()) {
            for (int n = q.size(); n; --n) {
                TreeNode* node = q.front();
                q.pop();
                if (vis.count(k - node->val)) {
                    return true;
                }
                vis.insert(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        return false;
    }
};

 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
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findTarget(root *TreeNode, k int) bool {
    q := []*TreeNode{root}
    vis := map[int]bool{}
    for len(q) > 0 {
        for n := len(q); n > 0; n-- {
            node := q[0]
            q = q[1:]
            if vis[k-node.Val] {
                return true
            }
            vis[node.Val] = true
            if node.Left != nil {
                q = append(q, node.Left)
            }
            if node.Right != nil {
                q = append(q, node.Right)
            }
        }
    }
    return false
}

 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
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function findTarget(root: TreeNode | null, k: number): boolean {
    const q = [root];
    const vis = new Set<number>();
    while (q.length) {
        for (let n = q.length; n; --n) {
            const { val, left, right } = q.shift();
            if (vis.has(k - val)) {
                return true;
            }
            vis.add(val);
            left && q.push(left);
            right && q.push(right);
        }
    }
    return false;
}

评论