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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:

*

* 1. 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]

* ]

*

*

* 2. 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]

* ]

*

*

* 3. 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 BinaryTreeVerticalOrderTraversal314 {

public List<List<Integer>> verticalOrder(TreeNode root) {

if (root == null) return new ArrayList<List<Integer>>();

Map<Integer, List<Integer>> map = new HashMap<>();

Queue<TreeNode> q = new LinkedList<>();

Map<TreeNode, Integer> vs = new HashMap<>();

q.offer(root);

vs.put(root, 0);

int min = 0;

while (!q.isEmpty()) {

TreeNode curr = q.poll();

int i = vs.get(curr);

if (curr.left != null) {

vs.put(curr.left, i-1);

q.offer(curr.left);

}

if (curr.right != null) {

vs.put(curr.right, i+1);

q.offer(curr.right);

}

map.computeIfAbsent(i, p -> new ArrayList<Integer>()).add(curr.val);

min = Math.min(min, i);

}

List<List<Integer>> res = new ArrayList<>();

while (map.containsKey(min)) {

res.add(map.get(min++));

}

return res;

}

public List<List<Integer>> verticalOrder2(TreeNode root) {

if (root == null) return new ArrayList<List<Integer>>();

Map<Integer, List<Integer>> map = new HashMap<>();

Queue<TreeNode> q = new LinkedList<>();

Queue<Integer> qi = new LinkedList<>();

q.add(root);

qi.add(0);

int min = 0;

while (!q.isEmpty()) {

TreeNode curr = q.remove();

Integer i = qi.remove();

if (curr.left != null) {

qi.add(i-1);

q.add(curr.left);

}

if (curr.right != null) {

qi.add(i+1);

q.add(curr.right);

}

if (!map.containsKey(i)) {

map.put(i, new ArrayList<Integer>());

}

map.get(i).add(curr.val);

// map.computeIfAbsent(i, p -> new ArrayList<Integer>()).add(curr.val);

min = Math.min(min, i);

}

List<List<Integer>> res = new ArrayList<>();

while (map.containsKey(min)) {

res.add(map.get(min++));

}

return res;

}

/**

* https://leetcode.com/problems/binary-tree-vertical-order-traversal/discuss/76401/5ms-Java-Clean-Solution

*/

public List<List<Integer>> verticalOrder3(TreeNode root) {

List<List<Integer>> cols = new ArrayList<>();

if (root == null) {

return cols;

}

int[] range = new int[] {0, 0};

getRange(root, range, 0);

for (int i = range[0]; i <= range[1]; i++) {

cols.add(new ArrayList<Integer>());

}

Queue<TreeNode> queue = new LinkedList<>();

Queue<Integer> colQueue = new LinkedList<>();

queue.add(root);

colQueue.add(-range[0]);

while (!queue.isEmpty()) {

TreeNode node = queue.poll();

int col = colQueue.poll();

cols.get(col).add(node.val);

if (node.left != null) {

queue.add(node.left);

colQueue.add(col - 1);

}

if (node.right != null) {

queue.add(node.right);

colQueue.add(col + 1);

}

}

return cols;

}

public void getRange(TreeNode root, int[] range, int col) {

if (root == null) {

return;

}

range[0] = Math.min(range[0], col);

range[1] = Math.max(range[1], col);

getRange(root.left, range, col - 1);

getRange(root.right, range, col + 1);

}

}