leetcode 501. Find Mode in Binary Search Tree

最新推荐文章于 2024-04-09 22:21:57 发布

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最新推荐文章于 2024-04-09 22:21:57 发布

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分类专栏： leetcode 文章标签： leetcode BST

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本文介绍了一种算法，用于在含有重复元素的二叉搜索树中寻找出现频率最高的元素（众数）。通过中序遍历的方式，可以有效地统计每个元素出现的次数，并找出众数。

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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).

最开始题目意思理解错了
这里要注意的是，如果有根的右子树中有与根值相同的节点，第一个这样的点不是直接连在根上，而是出现在子树的最左边。

由于是二叉搜索树，所以采用中序遍历得到的序列是从小到大的顺序。这样遍历到一个点时候，记录上一个点的值，来确定计数器的状态。从而更新结果list

public class Solution {
    int max = 0;
    Integer pre = null;
    int count = 1;
    public int[] findMode(TreeNode root) {
        if(root==null) return new int[0];

        ArrayList<Integer> resList = new ArrayList<>();
        sumChild(root, resList);
        int[] res = new int[resList.size()];
        for(int i=0; i< resList.size(); i++){
            res[i] = resList.get(i);
        }
        return res;
    }

    void sumChild(TreeNode root, ArrayList<Integer> list){
        if(root==null) return ;
        sumChild(root.left, list);
        if(pre!=null){
            if(root.val==pre){
                count++;
            }else{
                count = 1;
            }
        }

        if(count > max){
            list.clear();
            list.add(root.val);
            max = count;
        }else if(count==max){
            list.add(root.val);
        }
        pre = root.val;
        sumChild(root.right, list);
    }
}