FindModeInBinarySearchTreeOPT3 [source code]

public class FindModeInBinarySearchTreeOPT3 {
static
/******************************************************************************/
public class Solution {
    Integer prev = null;
    int count = 1;
    int max = 0;
    public int[] findMode(TreeNode root) {
        if (root == null) return new int[0];

        List<Integer> list = new ArrayList<>();
        traverse(root, list);

        int[] res = new int[list.size()];
        for (int i = 0; i < list.size(); ++i) res[i] = list.get(i);
        return res;
    }

    private void traverse(TreeNode root, List<Integer> list) {
        if (root == null) return;
        traverse(root.left, list);
        if (prev != null) {
            if (root.val == prev)
                count++;
            else
                count = 1;
        }
        if (count > max) {
            max = count;
            list.clear();
            list.add(root.val);
        } else if (count == max) {
            list.add(root.val);
        }
        prev = root.val;
        traverse(root.right, list);
    }
}
/******************************************************************************/

    public static void main(String[] args) {
        FindModeInBinarySearchTreeOPT3.Solution tester = new FindModeInBinarySearchTreeOPT3.Solution();
    }
}

这个是 discussion 最优解: 5(58);

没有想到居然这么快;
他这个空间严格来说用了一个 list, 所以实际上也是属于没有做到O(1)的空间. 如果opt2没有为了O(1)空间而专门走两个 pass, 感觉速度应该不比这个差. 因为 opt2的代码其实是属于非常清晰的了, 这个代码的思路基本也没有什么区别, 没有理由速度有很本质的区别;

Problem Description

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than or equal to the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

For example:
Given BST [1,null,2,2],
1
\
2
/
2
return [2].

Note: If a tree has more than one mode, you can return them in any order.

Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).

Difficulty:Easy
Category:Algorithms
Acceptance:37.96%
Contributor: Coder_1215
Companies
google
Related Topics
tree
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