987. Vertical Order Traversal of a Binary Tree
987. Vertical Order Traversal of a Binary Tree
Given a binary tree, return the vertical order traversal of its nodes values.
For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1)
and (X+1, Y-1).
Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes,
we report the values of the nodes in order from top to bottom (decreasing Y coordinates).
If two nodes have the same position, then the value of the node that is reported first is the value that is
smaller.
Return an list of non-empty reports in order of X coordinate. Every report will have a list of values of
nodes.
Example 1:
Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation:
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).
- Sorting
public List<List<Integer>> verticalTraversal(TreeNode root) {
PriorityQueue<Node> q = new PriorityQueue<>((a,b)->{
int c = Integer.compare(a.c, b.c);
if(c != 0) return c;
int r = Integer.compare(a.r, b.r);
if(r != 0) return r;
return Integer.compare(a.v, b.v);
});
traverse(root, q, 0, 0);
List<List<Integer>> result = new ArrayList<>();
Node pre = null;
ArrayList<Integer> line = null;
while(!q.isEmpty()) {
if(pre == null || pre.c != q.peek().c) {
if(pre != null) result.add(line);
line = new ArrayList<>();
}
pre = q.poll();
line.add(pre.v);
if(q.isEmpty()) {
result.add(line);
}
}
return result;
}
public void traverse(TreeNode root, PriorityQueue<Node> q, int r, int c) {
if(root == null) return;
q.offer(new Node(root.val, r, c));
traverse(root.left, q, r+1, c-1);
traverse(root.right, q, r+1, c+1);
}
class Node {
int v;
int r;
int c;
public Node(int v, int r, int c) {
this.v = v;
this.r = r;
this.c = c;
}
}