501-Find Mode in Binary Search Tree
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).
問題描述
給定一個有重複值的二叉排序樹, 找出其所有的模式(最頻繁的值)
假設二叉排序樹如下定義:
- 左子樹只包含小於等於節點值的節點
- 右子樹只包含大於等於節點值的節點
- 左右子樹也為二叉排序樹
問題分析
進行兩次中序遍歷, 第一次獲取模式個數(用來建立陣列)以及次數, 第二次往陣列裡面存值
解法
public class Solution {
private int currVal;
//當前值的個數
private int currCount = 0;
//當前最大個數
private int maxCount = 0;
//模式個數
private int modeCount = 0;
private int[] modes;
public int[] findMode(TreeNode root) {
inorder(root);
//注意這裡
modes = new int[modeCount];
//注意, 沒有重置maxCount
modeCount = 0;
currCount = 0;
inorder(root);
return modes;
}
private void handleValue(int val){
if(val != currVal){
currVal = val;
currCount = 0;
}
currCount++;
if(currCount > maxCount){
maxCount = currCount;
modeCount = 1;
}else if(currCount == maxCount){
if(modes != null) modes[modeCount] = currVal;
modeCount++;
}
}
private void inorder(TreeNode root){
if (root == null) return;
inorder(root.left);
handleValue(root.val);
inorder(root.right);
}
}
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