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
1 package uberOnsite; 2 3 import java.util.*; 4 5 public class PrintTree { 6 public static class TreeNode { 7 char val; 8 int col; 9 TreeNode left; 10 TreeNode right; 11 public TreeNode(char value) { 12 this.val = value; 13 } 14 } 15 16 static int colNum = 0; 17 18 public static List<String> print(TreeNode root) { 19 List<String> res = new ArrayList<String>(); 20 if (root == null) return res; 21 inorder(root); 22 levelOrder(root, res); 23 return res; 24 } 25 26 public static void inorder(TreeNode node) { 27 if (node == null) return; 28 inorder(node.left); 29 node.col = colNum; 30 colNum++; 31 inorder(node.right); 32 33 } 34 35 public static void levelOrder(TreeNode node, List<String> res) { 36 Queue<TreeNode> queue = new LinkedList<TreeNode>(); 37 queue.offer(node); 38 while (!queue.isEmpty()) { 39 StringBuilder line = new StringBuilder(); 40 HashMap<Integer, Character> lineMap = new HashMap<Integer, Character>(); 41 int maxCol = Integer.MIN_VALUE; 42 int size = queue.size(); 43 for (int i=0; i<size; i++) { 44 TreeNode cur = queue.poll(); 45 lineMap.put(cur.col, cur.val); 46 maxCol = Math.max(maxCol, cur.col); 47 if (cur.left != null) queue.offer(cur.left); 48 if (cur.right != null) queue.offer(cur.right); 49 } 50 for (int k=0; k<=maxCol; k++) { 51 if (lineMap.containsKey(k)) line.append(lineMap.get(k)); 52 else line.append(' '); 53 } 54 res.add(line.toString()); 55 } 56 } 57 58 /** 59 * @param args 60 */ 61 public static void main(String[] args) { 62 // TODO Auto-generated method stub 63 64 PrintTree sol = new PrintTree(); 65 66 TreeNode A = new TreeNode('a'); 67 TreeNode B = new TreeNode('b'); 68 TreeNode C = new TreeNode('c'); 69 TreeNode D = new TreeNode('d'); 70 TreeNode E = new TreeNode('e'); 71 TreeNode F = new TreeNode('f'); 72 TreeNode G = new TreeNode('g'); 73 TreeNode H = new TreeNode('h'); 74 TreeNode I = new TreeNode('i'); 75 TreeNode J = new TreeNode('j'); 76 TreeNode Z = new TreeNode('z'); 77 78 A.left = B; 79 A.right = C; 80 B.left = D; 81 C.left = E; 82 C.right = F; 83 D.left = Z; 84 D.right = G; 85 E.right = H; 86 F.left = I; 87 F.right = J; 88 89 List<String> res = print(A); 90 for (String each : res) { 91 System.out.println(each); 92 } 93 } 94 95 }
