653 Two Sum IV - Input is a BST
653. Two Sum IV - Input is a BST
1. Question
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.
Example 1:
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 9
Output:
True
Example 2:
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 28
Output:
False
2. Implementation
(1) DFS
class Solution {
public boolean findTarget(TreeNode root, int k) {
Set<Integer> set = new HashSet<>();
return findTarget(root, k, set);
}
public boolean findTarget(TreeNode node, int target, Set<Integer> set) {
if (node == null) {
return false;
}
if (set.contains(target - node.val)) {
return true;
}
set.add(node.val);
return findTarget(node.left, target, set) || findTarget(node.right, target, set);
}
}
(2) BFS
class Solution {
public boolean findTarget(TreeNode root, int k) {
Set<Integer> set = new HashSet<>();
if (root == null) {
return false;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode curNode = queue.remove();
if (set.contains(k - curNode.val)) {
return true;
}
set.add(curNode.val);
if (curNode.left != null) {
queue.add(curNode.left);
}
if (curNode.right != null) {
queue.add(curNode.right);
}
}
}
return false;
}
}
3. Time & Space Complexity
DFS: 时间复杂度: O(n), 空间复杂度: O(n)
BFS: 时间复杂度: O(n), 空间复杂度: O(w), w为一层中最多node的个数
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