Question

1214. Two Sum BSTs

中文文档

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.
# 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.
 * 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);
    nums[i].add(root.val);
    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.
 * 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;
}