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| 1 | +using System; |
| 2 | +using System.Collections.Generic; |
| 3 | +using System.Linq; |
| 4 | +using System.Text; |
| 5 | + |
| 6 | +namespace Tree.TreeLib |
| 7 | +{ |
| 8 | +// Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST. |
| 9 | + |
| 10 | +//Assume a BST is defined as follows: |
| 11 | + |
| 12 | +//The left subtree of a node contains only nodes with keys less than or equal to the node's key. |
| 13 | +//The right subtree of a node contains only nodes with keys greater than or equal to the node's key. |
| 14 | +//Both the left and right subtrees must also be binary search trees. |
| 15 | +//For example: |
| 16 | +//Given BST [1,null,2,2], |
| 17 | +// 1 |
| 18 | +// \ |
| 19 | +// 2 |
| 20 | +// / |
| 21 | +// 2 |
| 22 | +//return [2]. |
| 23 | + |
| 24 | + //求解二叉搜索树中出现次数最多的元素,答案的空间复杂度为 O(1)。 |
| 25 | + //二叉搜索树的中序遍历恰好为元素的从小到大排序,这样相等的元素一定是相邻的。 |
| 26 | + //了解了这个知识点,空间复杂度O(1)就可以做到了。 |
| 27 | + public class Findmodes |
| 28 | + { |
| 29 | + private int currentVal; |
| 30 | + private int currentCount = 0; |
| 31 | + private int maxCount = 0; |
| 32 | + private int modeCount = 0; |
| 33 | + private int[] modeArray; |
| 34 | + |
| 35 | + public int[] FindMode(TreeNode root) |
| 36 | + { |
| 37 | + preorder(root); //第一遍中序遍历找出出现次数最多的元素数,可能有多个最大 |
| 38 | + modeArray = new int[modeCount]; |
| 39 | + modeCount = 0; |
| 40 | + currentCount = 0; |
| 41 | + preorder(root); |
| 42 | + return modeArray; |
| 43 | + } |
| 44 | + |
| 45 | + /// <summary> |
| 46 | + /// 这种方法只适应于二叉搜索树条件下,查找元素出现的最多次数 |
| 47 | + /// </summary> |
| 48 | + /// <param name="val"></param> |
| 49 | + private void getModeValue(int val) |
| 50 | + { |
| 51 | + if (val != currentVal) |
| 52 | + { |
| 53 | + currentVal = val; |
| 54 | + currentCount = 0; |
| 55 | + } |
| 56 | + currentCount++; |
| 57 | + if (currentCount > maxCount) |
| 58 | + { |
| 59 | + maxCount = currentCount; |
| 60 | + modeCount = 1; |
| 61 | + } |
| 62 | + else if (currentCount == maxCount) |
| 63 | + { |
| 64 | + if (modeArray != null) //第二遍遍历后,对出现次数最多的元素 |
| 65 | + modeArray[modeCount] = currentVal; //依次赋值给modeArray |
| 66 | + modeCount++; |
| 67 | + } |
| 68 | + } |
| 69 | + |
| 70 | + /// <summary> |
| 71 | + /// 二叉搜索树,采取中序遍历对值处理 |
| 72 | + /// </summary> |
| 73 | + /// <param name="root"></param> |
| 74 | + private void preorder(TreeNode root) |
| 75 | + { |
| 76 | + if (root == null) |
| 77 | + return; |
| 78 | + preorder(root.left); |
| 79 | + getModeValue(root.val); |
| 80 | + preorder(root.right); |
| 81 | + } |
| 82 | + } |
| 83 | +} |
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