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]
]
思路
代码
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<List<Integer>> verticalOrder(TreeNode root) { 12 // corner 13 List<List<Integer>> results = new ArrayList<>(); 14 if (root == null) return results; 15 // init 16 int min = Integer.MAX_VALUE; 17 int max = Integer.MIN_VALUE; 18 Map<Integer, List<Integer>> map = new HashMap<>(); 19 Queue<Position> queue = new LinkedList<>(); 20 21 //use queue to bfs 22 queue.add(new Position(root, 0)); 23 while (!queue.isEmpty()) { 24 Position position = queue.remove(); 25 min = Math.min(min, position.column); 26 max = Math.max(max, position.column); 27 List<Integer> list = map.get(position.column); 28 if (list == null) { 29 list = new ArrayList<>(); 30 map.put(position.column, list); 31 } 32 list.add(position.node.val); 33 if (position.node.left != null) queue.add(new Position(position.node.left, position.column-1)); 34 if (position.node.right != null) queue.add(new Position(position.node.right, position.column+1)); 35 } 36 for(int i = min; i<= max; i++) { 37 List<Integer> list = map.get(i); 38 if (list != null) results.add(list); 39 } 40 return results; 41 } 42 } 43 44 class Position { 45 TreeNode node; 46 int column; 47 Position(TreeNode node, int column) { 48 this.node = node; 49 this.column = column; 50 } 51 }
