LeetCode Find Mode in Binary Search Tree

原题链接在这里：https://leetcode.com/problems/find-mode-in-binary-search-tree/#/description

题目：

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).

题解：

两遍Binary Tree Inorder Traversal.

第一遍找出有几个mode. 建立res array, 并保留了mode duplicate次数是多少. 第二遍当duplicate次数是mode 的duplicate次数时就加入res中.

Time Complexity: O(n).

Space: O(n), 每个节点都不同, res的size就是O(n). stack space O(logn). 如果利用Binary Tree Inorder Traversal中的Morris Traversal方法可以不用stack space.

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     int curVal;
     int curCount;
     int maxCount;
     int modeNumber;
     int [] res;

     public int[] findMode(TreeNode root) {
         inorderTraversal(root);
         res = new int[modeNumber];
         curCount = 0;
         modeNumber = 0;
         inorderTraversal(root);
         return res;
     }

     private void inorderTraversal(TreeNode root){
         if(root == null){
             return;
         }

         inorderTraversal(root.left);
         handleCurrentNodeValue(root.val);
         inorderTraversal(root.right);
     }

     private void handleCurrentNodeValue(int val){
         if(val != curVal){
             curVal = val;
             curCount = 0;
         }
         curCount++;

         if(curCount > maxCount){
             maxCount = curCount;
             modeNumber = 1;
         }else if(curCount == maxCount){
             if(res != null){
                 res[modeNumber] = curVal;
             }
             modeNumber++;
         }
     }
 }