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);
         }
     }

 }