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Solution2.java
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> verticalOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<Pair<TreeNode, Integer>> q = new ArrayDeque<>();
q.offer(new Pair<>(root, 0));
TreeMap<Integer, List<Integer>> d = new TreeMap<>();
while (!q.isEmpty()) {
for (int n = q.size(); n > 0; --n) {
var p = q.pollFirst();
root = p.getKey();
int offset = p.getValue();
d.computeIfAbsent(offset, k -> new ArrayList()).add(root.val);
if (root.left != null) {
q.offer(new Pair<>(root.left, offset - 1));
}
if (root.right != null) {
q.offer(new Pair<>(root.right, offset + 1));
}
}
}
return new ArrayList<>(d.values());
}
}