314. Binary Tree Vertical Order Traversal

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] ]

Solution

/**
 * 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) {
        if (root == null) return new ArrayList<>();
        Map<Integer, List<Integer>> map = new HashMap<>();


        Queue<Result> queue = new LinkedList<>();
        queue.add(new Result(root, 0));

        while (!queue.isEmpty()) {
            Result cur = queue.poll();
            TreeNode node = cur.node;

            List<Integer> curList = new ArrayList<>();
            curList.add(cur.node.val);
            map.merge(cur.index, curList, (a, b) -> {
                a.addAll(b);
                return a;
            });

            if (node.left != null) {
                queue.add(new Result(node.left, cur.index-1));
            }

            if (node.right != null) {
                queue.add(new Result(node.right, cur.index+1));
            }
        }

        List<Integer> keys = map.keySet().stream().sorted().collect(Collectors.toList());
        List<List<Integer>> result = new ArrayList<>();
        for (Integer key : keys) {
            result.add(map.get(key));
        }

        return result;

    }

    private class Result {
        TreeNode node;
        int index;

        public Result(TreeNode node, int index) {
            this.node = node;
            this.index = index;
        }
    }
}