精通算法系列-二叉树问题

原题为:

// 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; }
 * }
 */

答案如下：

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;

    }

}