leetcode-501-Find-Mode-in-Binary-Search-Tree

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描述

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],

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2
3
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5
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).

分析

求二分搜索树中的众数，因为二分搜索树的节点值是有序的，所以遍历一遍就够了，不需要使用哈希列表，要注意的是这道题有可能出现多个众数。

解决方案1(Java)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    Integer prevNode = null;
    int count = 1;
    int nowMax = 0;
    
    public int[] findMode(TreeNode root) {
        if (root == null) {
            return new int[0];
        }
        List<Integer> list = new ArrayList<>();
        inOrder(root, list);
        int[] result = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            result[i] = list.get(i);
        }
        return result;
    }
    
    private void inOrder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        inOrder(root.left, list);
        if (prevNode != null) {
            if (prevNode == root.val) {
                count++;
            } else {
                count = 1;
            }
        }
        
        if (count > nowMax) {
            nowMax = count;
            list.clear();
            list.add(root.val);
        } else if (count == nowMax) {
            list.add(root.val);
        }
        prevNode = root.val;
        inOrder(root.right, list);
    }
}

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