Binary Tree Vertical Order Traversal
Source
- leetcode: 314
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.
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]
]
Given binary tree [3,9,20,4,5,2,7],
_3_
/ \
9 20
/ \ / \
4 5 2 7
return its vertical order traversal as:
[
[4],
[9],
[3,5,2],
[20],
[7]
]
题解
he following solution takes 5ms.
BFS, put node, col into queue at the same time Every left child access col - 1 while right child col + 1 This maps node into different col buckets Get col boundary min and max on the fly Retrieve result from cols Note that TreeMap version takes 9ms.
Here is an example of [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]. Notice that every child access changes one column bucket id. So 12 actually goes ahead of 11.
Java
public List<List<Integer>> verticalOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
Map<Integer, ArrayList<Integer>> map = new HashMap<>();
Queue<TreeNode> q = new LinkedList<>();
Queue<Integer> cols = new LinkedList<>();
q.add(root);
cols.add(0);
int min = 0, max = 0;
while(!q.isEmpty()) {
TreeNode node = q.poll();
int col = cols.poll();
if(!map.containsKey(col)) map.put(col, new ArrayList<Integer>());
map.get(col).add(node.val);
if(node.left != null) {
q.add(node.left);
cols.add(col - 1);
if(col <= min) min = col - 1;
}
if(node.right != null) {
q.add(node.right);
cols.add(col + 1);
if(col >= max) max = col + 1;
}
}
for(int i = min; i <= max; i++) {
res.add(map.get(i));
}
return res;
}