314.Binary Tree Vertical Order traversal

Source: Internet
Author: User


Given a binary tree, return the vertical order Traversal of its nodes ' values. (ie, from top to bottom, and column by column).

If the nodes is in the same row and column, the order should is from left to right.

Given binary Tree [3,9,20,null,null,15,7] ,

    3   /   9    /   7

Return its vertical order traversal as:

[  9],  [3,15],  [+],  [7]]

Given binary Tree [3,9,20,4,5,2,7] ,

    _3_   /     9    /\   /4   5 2   7

Return its vertical order traversal as:

[  4],  [9],  [3,5,2],  [+],  [7]]

Links: http://leetcode.com/problems/binary-tree-vertical-order-traversal/


Binary Tree Vertical order traversal. The meaning of this question is very simple, but the example is not good enough, if the second example above 5 also has the right subtree, and 20 in a column. In general it is assumed that a node column is I, then its left subtree column is i-1, right subtree column is i + 1. We can use the decorator mode to build a treecolumnnode that contains a TreeNode, and a column value, and then use the level order traversal to calculate it. Use a hashmap to save the column value and the point of the same value when calculating. Also set a min column value and a max column value to make it easy to finally get the value output in HashMap from small to large order.

Time Complexity-o (n), Space complexity-o (n)

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode rig Ht * TreeNode (int x) {val = x;} }*/ Public classSolution {Private classtreecolumnnode{ PublicTreeNode TreeNode; intCol;  PublicTreecolumnnode (TreeNode node,intCol) {             This. TreeNode =node;  This. Col =Col; }    }         PublicList<list<integer>>Verticalorder (TreeNode root) {List<List<Integer>> res =NewArraylist<>(); if(Root = =NULL) {            returnRes; } Queue<TreeColumnNode> queue =NewLinkedlist<>(); Map<integer, list<integer>> map =NewHashmap<>(); Queue.offer (NewTreecolumnnode (Root, 0)); intCurlevel = 1; intNextlevel = 0; intMin = 0; intMax = 0;  while(!Queue.isempty ()) {treecolumnnode node=Queue.poll (); if(Map.containskey (Node.col)) {Map.get (Node.col). Add (Node.treeNode.val); } Else{map.put (Node.col,NewArraylist<integer>(Arrays.aslist (Node.treeNode.val))); } curlevel--; if(Node.treeNode.left! =NULL) {Queue.offer (NewTreecolumnnode (Node.treeNode.left, NODE.COL-1)); Nextlevel++; Min= Math.min (node.col-1, Min); }            if(Node.treeNode.right! =NULL) {Queue.offer (NewTreecolumnnode (Node.treeNode.right, Node.col + 1)); Nextlevel++; Max= Math.max (Node.col + 1, Max); }            if(Curlevel = = 0) {Curlevel=Nextlevel; Nextlevel= 0; }        }                 for(inti = min; I <= Max; i++) {Res.add (Map.get (i)); }                returnRes; }}









314.Binary Tree Vertical Order traversal

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