1214. Two Sum BSTs

Description

Given the roots of two binary search trees, root1 and root2, return true if and only if there is a node in the first tree and a node in the second tree whose values sum up to a given integer target.

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3], target = 5
Output: true
Explanation: 2 and 3 sum up to 5.


Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2], target = 18
Output: false


Constraints:

• The number of nodes in each tree is in the range [1, 5000].
• -109 <= Node.val, target <= 109

Solutions

Solution 1: In-order Traversal + Two Pointers

We perform in-order traversals on the two trees separately, obtaining two sorted arrays $nums[0]$ and $nums[1]$. Then we use a two-pointer method to determine whether there exist two numbers whose sum equals the target value. The two-pointer method is as follows:

Initialize two pointers $i$ and $j$, pointing to the left boundary of array $nums[0]$ and the right boundary of array $nums[1]$ respectively;

Each time, compare the sum $x = nums[0][i] + nums[1][j]$ with the target value. If $x = target$, return true; otherwise, if $x \lt target$, move $i$ one step to the right; otherwise, if $x \gt target$, move $j$ one step to the left.

The time complexity is $O(m + n)$, and the space complexity is $O(m + n)$. Here, $m$ and $n$ are the number of nodes in the two trees respectively.

• /**
* 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 List<Integer>[] nums = new List[2];

public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
Arrays.setAll(nums, k -> new ArrayList<>());
dfs(root1, 0);
dfs(root2, 1);
int i = 0, j = nums[1].size() - 1;
while (i < nums[0].size() && j >= 0) {
int x = nums[0].get(i) + nums[1].get(j);
if (x == target) {
return true;
}
if (x < target) {
++i;
} else {
--j;
}
}
return false;
}

private void dfs(TreeNode root, int i) {
if (root == null) {
return;
}
dfs(root.left, i);
dfs(root.right, i);
}
}

• /**
* 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 twoSumBSTs(TreeNode* root1, TreeNode* root2, int target) {
vector<int> nums[2];
function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int i) {
if (!root) {
return;
}
dfs(root->left, i);
nums[i].push_back(root->val);
dfs(root->right, i);
};
dfs(root1, 0);
dfs(root2, 1);
int i = 0, j = nums[1].size() - 1;
while (i < nums[0].size() && j >= 0) {
int x = nums[0][i] + nums[1][j];
if (x == target) {
return true;
}
if (x < target) {
++i;
} else {
--j;
}
}
return false;
}
};

• # 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 twoSumBSTs(
self, root1: Optional[TreeNode], root2: Optional[TreeNode], target: int
) -> bool:
def dfs(root: Optional[TreeNode], i: int):
if root is None:
return
dfs(root.left, i)
nums[i].append(root.val)
dfs(root.right, i)

nums = [[], []]
dfs(root1, 0)
dfs(root2, 1)
i, j = 0, len(nums[1]) - 1
while i < len(nums[0]) and ~j:
x = nums[0][i] + nums[1][j]
if x == target:
return True
if x < target:
i += 1
else:
j -= 1
return False


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
nums := [2][]int{}
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, i int) {
if root == nil {
return
}
dfs(root.Left, i)
nums[i] = append(nums[i], root.Val)
dfs(root.Right, i)
}
dfs(root1, 0)
dfs(root2, 1)
i, j := 0, len(nums[1])-1
for i < len(nums[0]) && j >= 0 {
x := nums[0][i] + nums[1][j]
if x == target {
return true
}
if x < target {
i++
} else {
j--
}
}
return false
}

• /**
* 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 twoSumBSTs(root1: TreeNode | null, root2: TreeNode | null, target: number): boolean {
const nums: number[][] = Array(2)
.fill(0)
.map(() => []);
const dfs = (root: TreeNode | null, i: number) => {
if (!root) {
return;
}
dfs(root.left, i);
nums[i].push(root.val);
dfs(root.right, i);
};
dfs(root1, 0);
dfs(root2, 1);
let i = 0;
let j = nums[1].length - 1;
while (i < nums[0].length && j >= 0) {
const x = nums[0][i] + nums[1][j];
if (x === target) {
return true;
}
if (x < target) {
++i;
} else {
--j;
}
}
return false;
}