314. Binary Tree Vertical Order Traversal
package LeetCode_314 import java.util.* import kotlin.collections.ArrayList import kotlin.collections.HashMap /** * 314. Binary Tree Vertical Order Traversal * (Prime) * Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column). If two nodes are in the same row and column, the order should be from left to right. Examples 1: Input: [3,9,20,null,null,15,7] 3 /\ / \ 9 20 /\ / \ 15 7 Output: [ [9], [3,15], [20], [7] ] * */ class TreeNode(var `val`:Int){ var left: TreeNode?=null var right: TreeNode?=null } class Node(root: TreeNode?, hd:Int){ var root:TreeNode?=null var horizontalDistance = 0 init{ this.root = root this.horizontalDistance = hd } } class Solution { /* * solution: BFS+HashMap, BFS to traverse all node, and HashMap to save elements of each coordinate * Time complexity:O(n), Space complexity:O(n) * */ fun verticalOrder(root: TreeNode?): List<List<Int>>{ val result = ArrayList<ArrayList<Int>>() if (root==null){ return result } //for the range of result var minHD = Int.MAX_VALUE var maxHD = Int.MIN_VALUE //key is coordinate, value is elements of this coordinate val map = HashMap<Int,ArrayList<Int>>() val queue = LinkedList<Node>() //the coordinate root is 0, root.left should be -1, root.right should be +1 val starting = Node(root,0) queue.offer(starting) while (queue.isNotEmpty()){ val cur = queue.pop() if (cur != null) { map.putIfAbsent(cur.horizontalDistance, ArrayList()) minHD = Math.min(minHD, cur.horizontalDistance) maxHD = Math.max(maxHD, cur.horizontalDistance) if (cur.root != null) { map.get(cur.horizontalDistance)!!.add(cur.root!!.`val`) if (cur.root!!.left!=null){ queue.offer(Node(cur.root!!.left,cur.horizontalDistance - 1)) } if (cur.root!!.right!=null){ queue.offer(Node(cur.root!!.right,cur.horizontalDistance + 1)) } } } } for (i in minHD..maxHD){ if (map.get(i)!=null){ result.add(map.get(i)!!) } } return result } }
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2019-08-04 Min Heap in Kotlin