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binaryTreeVerticalOrderTraversal.java
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// 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 [3,9,8,4,0,1,7],
// 3
// /\
// / \
// 9 8
// /\ /\
// / \/ \
// 4 01 7
// return its vertical order traversal as:
// [
// [4],
// [9],
// [3,0,1],
// [8],
// [7]
// ]
// 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]
// ]
/**
* 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>> result = new ArrayList<>();
if(root == null) return result;
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;
int 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);
min = Math.min(min, col - 1);
}
if(node.right != null) {
q.add(node.right);
cols.add(col + 1);
max = Math.max(max, col + 1);
}
}
for(int i = min; i <= max; i++) {
result.add(map.get(i));
}
return result;
}
}