314. Binary Tree Vertical Order Traversal (Medium)
Given a binary tree, return thevertical ordertraversal 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.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]
]
2.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]
]
3.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 1: BFS + HashMap O(n); O(w + n)
public List<List<Integer>> verticalOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Map<Integer, List<Integer>> map = new HashMap<>(); //O(n)
Queue<TreeNode> queue = new LinkedList<>(); //O(w)
Queue<Integer> cols = new LinkedList<>();
queue.offer(root);
cols.offer(0);
int min = 0;
int max = 0;
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
int col = cols.poll();
if (!map.containsKey(col)) {
map.put(col, new ArrayList<>());
}
map.get(col).add(node.val);
if (node.left != null) {
queue.offer(node.left);
cols.offer(col - 1);
min = Math.min(min, col - 1);
}
if (node.right != null) {
queue.offer(node.right);
cols.offer(col + 1);
max = Math.max(max, col + 1);
}
}
for (int i = min; i < max; i++) {
res.add(map.get(i));
}
return res;
}