leetcode 653. Two Sum IV

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   7Target = 9Output: True

Example 2:

Input:     5   / \  3   6 / \   \2   4   7Target = 28Output: False

简单题一个。

public boolean findTarget(TreeNode root, int k) {HashSet<Integer> hashSet=new HashSet<Integer>();return find(root, k, hashSet);}public boolean find(TreeNode node,int k,HashSet<Integer> hashSet){if(node==null){return false;}if(hashSet.contains(k-node.val)){return true;}hashSet.add(node.val);return find(node.left, k, hashSet)||find(node.right, k, hashSet);}

有大神发现，这是一个BST，就可以用BST的性质来做：

1. 中根遍历 BST，能得到一个从小到大排序的 list
2. 在 排好序的 list 中来解决 Two Sum 问题

public class Solution {    public boolean findTarget(TreeNode root, int k) {        if (root == null) return false;                List<Integer> list = new ArrayList<>();        inOrder(root, list);                int i = 0, j = list.size() - 1;        while (i < j) {            int sum = list.get(i) + list.get(j);            if (sum == k) return true;            if (sum < k) {                i++;            }            else {                j--;            }        }                return false;    }        private void inOrder(TreeNode root, List<Integer> list) {        if (root == null) return;                inOrder(root.left, list);        list.add(root.val);        inOrder(root.right, list);    }}

这道题有solutions：https://leetcode.com/problems/two-sum-iv-input-is-a-bst/solution/

Solution

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Approach #1 Using HashSet[Accepted]

Java

public class Solution {    public boolean findTarget(TreeNode root, int k) {        Set < Integer > set = new HashSet();        return find(root, k, set);    }    public boolean find(TreeNode root, int k, Set < Integer > set) {        if (root == null)            return false;        if (set.contains(k - root.val))            return true;        set.add(root.val);        return find(root.left, k, set) || find(root.right, k, set);    }}

Complexity Analysis

• Time complexity : $O(n)$. The entire tree is traversed only once in the worst case. Here, $n$ refers to the number of nodes in the given tree.

• Space complexity : $O(n)$. The size of the $set$ can grow upto $n$ in the worst case.

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Approach #2 Using BFS and HashSet [Accepted]

Java

public class Solution {    public boolean findTarget(TreeNode root, int k) {        Set < Integer > set = new HashSet();        Queue < TreeNode > queue = new LinkedList();        queue.add(root);        while (!queue.isEmpty()) {            if (queue.peek() != null) {                TreeNode node = queue.remove();                if (set.contains(k - node.val))                    return true;                set.add(node.val);                queue.add(node.right);                queue.add(node.left);            } else                queue.remove();        }        return false;    }}

Complexity Analysis

• Time complexity : $O(n)$. We need to traverse over the whole tree once in the worst case. Here, $n$ refers to the number of nodes in the given tree.

• Space complexity : $O(n)$. The size of the $set$ can grow atmost upto $n$.

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Approach #3 Using BST [Accepted]

Java

public class Solution {    public boolean findTarget(TreeNode root, int k) {        List < Integer > list = new ArrayList();        inorder(root, list);        int l = 0, r = list.size() - 1;        while (l < r) {            int sum = list.get(l) + list.get(r);            if (sum == k)                return true;            if (sum < k)                l++;            else                r--;        }        return false;    }    public void inorder(TreeNode root, List < Integer > list) {        if (root == null)            return;        inorder(root.left, list);        list.add(root.val);        inorder(root.right, list);    }}

Complexity Analysis

• Time complexity : $O(n)$. We need to traverse over the whole tree once to do the inorder traversal. Here, $n$ refers to the number of nodes in the given tree.

• Space complexity : $O(n)$. The sorted $list$ will contain $n$ elements.