653. Two Sum IV - Input is a BST

问题
Given the root of a Binary Search Tree and a target number k, return true if there exist two elements in the BST such that their sum is equal to the given target.

DFS搜索,每次递归时检查HashSet中是否有当前节点的值。如没有则将目标值减去当前节点的值加入HashSet。如有则返回true。
递归左侧节点和右侧节点,并返回二者的或运算。

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/**
* 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 {
HashSet<Integer> set;
public boolean findTarget(TreeNode root, int k) {
set = new HashSet();
return dfs(root,k);
}

private boolean dfs(TreeNode root, int k){
if(root == null){
return false;
}
if(set.contains(root.val)){
return true;
}
set.add(k - root.val);

return dfs(root.left,k) || dfs(root.right,k);
}
}
Author

Xander

Posted on

2022-04-16

Updated on

2022-04-16

Licensed under

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