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Problem
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.
Example
Examples 1:
Input: [3,9,20,null,null,15,7]
3
/
/
9 20
/
/
15 7
Output:
[
[9],
[3,15],
[20],
[7]
]Examples 2:
Input: [3,9,8,4,0,1,7]
3
/
/
9 8
/ /
/ /
4 01 7
Output:
[
[4],
[9],
[3,0,1],
[8],
[7]
]Examples 3:
Input: [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
Output:
[
[4],
[9,5],
[3,0,1],
[8,2],
[7]
]
Solution
class Solution {
public List<List<Integer>> verticalOrder(TreeNode root) {
//result array and edge case
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
//create map and queues
Map<Integer, List<Integer>> map = new HashMap<>();
Queue<TreeNode> nodes = new LinkedList<>();
Queue<Integer> indices = new LinkedList<>();
//initialize the queues
nodes.offer(root);
indices.offer(0);
//bfs to traverse all nodes
while (!nodes.isEmpty()) {
TreeNode node = nodes.poll();
Integer index = indices.poll();
map.putIfAbsent(index, new ArrayList<Integer>());
map.get(index).add(node.val);
if (node.left != null) {
nodes.offer(node.left);
indices.offer(index-1);
}
if (node.right != null) {
nodes.offer(node.right);
indices.offer(index+1);
}
}
//well, the list needs to be ordered
for (int i = Collections.min(map.keySet()); i <= Collections.max(map.keySet()); i++) {
res.add(map.get(i));
}
return res;
}
}
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