653. Two Sum IV - Input is a BST

最新推荐文章于 2020-02-25 09:07:10 发布

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博客围绕在二叉搜索树（BST）中查找是否存在两元素和等于给定目标值的问题展开。题目要求判断BST里是否有满足条件的元素。方法思路提到遍历数组查找，指出中序遍历比先序遍历更优，还给出了基于先序和中序的两种方法。

题目描述

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.

在这里插入图片描述

方法思路

这道题目相对简单，只要遍历了数组并进行查找即可，相比较而言，中序遍历要比先序遍历更好一点，因为前者利用了BST的结构特性。

Approach1: based on preorder

class Solution {
    //Runtime: 13 ms, faster than 78.49% 
    //Memory Usage: 46.4 MB, less than 24.87%
    List<Integer> list;
    public boolean findTarget(TreeNode root, int k) {
        list = new ArrayList<>();
        dfs(root);
        for(int i = 0; i < list.size(); i++){
            //要考虑到k = 2 * list.get(i)的情况
            if(list.contains(k - list.get(i)) && k != 2 * list.get(i))
                return true;
        }
        return false;
    }
    public void dfs(TreeNode root){
        if(root == null) return;
        list.add(root.val);
        dfs(root.left);
        dfs(root.right);
        return;
    }
}

Approach2: based on inorder

public class Solution {
    //Runtime: 12 ms, faster than 91.82%
    //Memory Usage: 44.8 MB, less than 33.89%
    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);
    }
}