U面经Prepare: Print Binary Tree With No Two Nodes Share The Same Column

Give a binary tree, elegantly print it so that no two tree nodes share the same column. 

Requirement: left child should appear on the left column of root, and right child should appear on the right of root.

Example: 
    a
   b   c
 d   e   f
z g   h i j

这道题若能发现inorder traversal each node的顺序其实就是column number递增的顺序，那么就成功了一大半

维护一个global variable，colNum, 做inorder traversal

然后level order 一层一层打印出来

package uberOnsite;

import java.util.*;

public class PrintTree {
    public static class TreeNode {
        char val;
        int col;
        TreeNode left;
        TreeNode right;
        public TreeNode(char value) {
            this.val = value;
        }
    }
    
    static int colNum = 0;
    
    public static List<String> print(TreeNode root) {
        List<String> res = new ArrayList<String>();
        if (root == null) return res;
        inorder(root);
        levelOrder(root, res);
        return res;
    }
    
    public static void inorder(TreeNode node) {
        if (node == null) return;
        inorder(node.left);
        node.col = colNum;
        colNum++;
        inorder(node.right);
        
    }
    
    public static void levelOrder(TreeNode node, List<String> res) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(node);
        while (!queue.isEmpty()) {
            StringBuilder line = new StringBuilder();
            HashMap<Integer, Character> lineMap = new HashMap<Integer, Character>();
            int maxCol = Integer.MIN_VALUE;
            int size = queue.size();
            for (int i=0; i<size; i++) {
                TreeNode cur = queue.poll();
                lineMap.put(cur.col, cur.val);
                maxCol = Math.max(maxCol, cur.col);
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            for (int k=0; k<=maxCol; k++) {
                if (lineMap.containsKey(k)) line.append(lineMap.get(k));
                else line.append(' ');
            }
            res.add(line.toString());
        }
    }
    
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        
        PrintTree sol = new PrintTree();
        
        TreeNode A = new TreeNode('a');
        TreeNode B = new TreeNode('b');
        TreeNode C = new TreeNode('c');
        TreeNode D = new TreeNode('d');
        TreeNode E = new TreeNode('e');
        TreeNode F = new TreeNode('f');
        TreeNode G = new TreeNode('g');
        TreeNode H = new TreeNode('h');
        TreeNode I = new TreeNode('i');
        TreeNode J = new TreeNode('j');
        TreeNode Z = new TreeNode('z');
        
        A.left = B;
        A.right = C;
        B.left = D;
        C.left = E;
        C.right = F;
        D.left = Z;
        D.right = G;
        E.right = H;
        F.left = I;
        F.right = J;
        
        List<String> res = print(A);
        for (String each : res) {
            System.out.println(each);
        }
    }

}