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Top view of a binary tree is the set of nodes visible when the tree is viewed from the top. Given a binary tree, print the top view of it. The output nodes can be printed in any order. Expected time complexity is O(n)
A node x is there in output if x is the topmost node at its horizontal distance. Horizontal distance of left child of a node x is equal to horizontal distance of x minus 1, and that of right child is horizontal distance of x plus 1.
1
/ 2 3
/ \ / 4 5 6 7
Top view of the above binary tree is
4 2 1 3 7
1
/ 2 3
\
4
5
6
Top view of the above binary tree is
2 1 3 6
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The idea is to do something similar to vertical Order Traversal. Like vertical Order Traversal, we need to nodes of same horizontal distance together. We do a level order traversal so that the topmost node at a horizontal node is visited before any other node of same horizontal distance below it. Hashing is used to check if a node at given horizontal distance is seen or not.
1 /Print a Binary Tree in Vertical Order 2 static int min; 3 static int max; 4 static int[] output; 5 6 public class Item{ 7 public Integer dis; 8 public TreeNode root; 9 public Item(Integer dis, TreeNode root){ 10 this.root = root; 11 this.dis = dis; 12 } 13 } 14 static int min; 15 static int max; 16 static int[] output; 17 18 public static void findMinMax(TreeNode root, Integer dis){ 19 if(root == null) return; 20 else{ 21 min = Math.min(dis, min); 22 max = Math.max(dis, max); 23 } 24 findMinMax(root.left, dis - 1); 25 findMinMax(root.right, dis + 1); 26 } 27 28 public static void levelOrder(TreeNode root){ 29 LinkedList<Item> queue = new LinkedList<Item>(); 30 queue.add(new Item(0, root)); 31 while(!queue.isEmpty()){ 32 Item tmp = queue.poll(); 33 // if(output[tmp.dis - min] == 0){ 34 output[tmp.dis - min] = tmp.root.val; 35 // } 36 if(tmp.root.left != null) queue.add(new Item(tmp.dis - 1, tmp.root.left)); 37 if(tmp.root.right != null) queue.add(new Item(tmp.dis + 1, tmp.root.right)); 38 } 39 } 40 41 public static int[] verticalOrderTraveralBT(TreeNode root){ 42 min = 0; max = 0; 43 findMinMax(root, 0); 44 int len = max - min + 1; 45 output = new int[len]; 46 levelOrder(root); 47 return output; 48 } 49 50 51 public static void main(String[] args) { 52 // int[] p = new int[]{10, 20, 30, 40, 30}; 53 // System.out.println(MatrixChainMultiplication(p)); 54 55 56 TreeNode root = new TreeNode(1); 57 root.left = new TreeNode(2); 58 root.right = new TreeNode(3); 59 root.left.left = new TreeNode(4); 60 root.left.right = new TreeNode(5); 61 root.right.left = new TreeNode(6); 62 root.right.right = new TreeNode(7); 63 root.right.left.right = new TreeNode(8); 64 root.right.right.right = new TreeNode(9); 65 66 /* Create following Binary Tree 67 1 68 / 69 2 3 70 71 4 72 73 5 74 75 6*/ 76 // TreeNode root = new TreeNode(1); 77 // root.left = new TreeNode(2); 78 // root.right = new TreeNode(3); 79 // root.left.right = new TreeNode(4); 80 // root.left.right.right = new TreeNode(5); 81 // root.left.right.right.right = new TreeNode(6); 82 int[] result = verticalOrderTraveralBT(root); 83 System.out.println(result); 84 }
如果是top view 就把 if(output[tmp.dis - min] == 0){ uncomment
Print Nodes in Top View of Binary Tree
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原文地址：http://www.cnblogs.com/reynold-lei/p/4338683.html
