Binary Tree Vertical Order Traversal
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 fromleft 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]
]
图例
分析
代码
/**
* 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) {
List<List<Integer>> results = new ArrayList<>(); //
if (root == null) return results;
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
Map<Integer, List<Integer>> map = new HashMap<>();
LinkedList<Position> queue = new LinkedList<>();
queue.add(new Position(root, 0));
while (!queue.isEmpty()) {
Position position = queue.remove();·
min = Math.min(min, position.column);
max = Math.max(max, position.column);
List<Integer> list = map.get(position.column);
if (list == null) {
list = new ArrayList<>();
map.put(position.column, list);
}
list.add(position.node.val);
if (position.node.left != null) queue.add(new Position(position.node.left, position.column-1));
if (position.node.right != null) queue.add(new Position(position.node.right, position.column+1));
}
for(int i=min; i<=max; i++) {
List<Integer> list = map.get(i);
if (list != null) results.add(list);
}
return results;
}
}
// write a wrapper class
class Position {
TreeNode node;
int column;
Position(TreeNode node, int column) {
this.node = node;
this.column = column;
}
}