314. Binary Tree Vertical Order Traversal

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:

Given binary tree [3,9,20,null,null,15,7], 3 /\ / \ 9 20 /\ / \ 15 7

return its vertical order traversal as:

[ [9], [3,15], [20], [7] ] Given binary tree [3,9,8,4,0,1,7], 3 /\ / \ 9 8 /\ /\ / \/ \ 4 01 7

return its vertical order traversal as:

[ [4], [9], [3,0,1], [8], [7] ] Given binary tree [3,9,8,4,0,1,7,null,null,null,2,5] (0's right child is 2 and 1's left child is 5), 3 /\ / \ 9 8 /\ /\ / \/ \ 4 01 7 /\ / \ 5 2

return its vertical order traversal as:

[ [4], [9,5], [3,0,1], [8,2], [7] ] 二叉树的竖序遍历，代码如下：

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> verticalOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<List<Integer>>(); if (root == null) { return res; } HashMap<Integer,List<Integer>> map = new HashMap<Integer,List<Integer>>(); int min = 0, max = 0; Queue<TreeNode> q = new LinkedList<>(); Queue<Integer> cols = new LinkedList<>(); q.add(root); cols.add(0); while (!q.isEmpty()) { TreeNode node = q.poll(); int col = cols.poll(); if (!map.containsKey(col)) { map.put(col, new ArrayList<Integer>()); } map.get(col).add(node.val); if (node.left != null) { q.add(node.left); cols.add(col - 1); min = Math.min(min, col - 1); } if (node.right != null) { q.add(node.right); cols.add(col + 1); max = Math.max(max, col + 1); } } for (int i = min; i <= max; i ++) { res.add(map.get(i)); } return res; } }