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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:
[3,9,20,null,null,15,7], 3
/ / 9 20
/ / 15 7
return its vertical order traversal as:
[ [9], [3,15], [20], [7] ]
[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] ]
[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] ]
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<List<Integer>> verticalOrder(TreeNode root) { 12 List<List<Integer>> resList = new ArrayList<List<Integer>>(); 13 if (root==null) return resList; 14 15 List<TreeNode> nodeList = new ArrayList<TreeNode>(); 16 List<Integer> indexList = new ArrayList<Integer>(); 17 int cur = 0, min = 0, max=0; 18 nodeList.add(root); 19 indexList.add(0); 20 21 while (cur<nodeList.size()){ 22 TreeNode curNode = nodeList.get(cur); 23 int curIndex = indexList.get(cur); 24 if (curIndex < min) min = curIndex; 25 if (curIndex > max) max = curIndex; 26 27 if (curNode.left!=null){ 28 nodeList.add(curNode.left); 29 indexList.add(curIndex-1); 30 } 31 if (curNode.right!=null){ 32 nodeList.add(curNode.right); 33 indexList.add(curIndex+1); 34 } 35 cur++; 36 } 37 38 for (int i=min;i<=max;i++) resList.add(new LinkedList<Integer>()); 39 40 for (int i=0;i<nodeList.size();i++){ 41 resList.get(indexList.get(i)-min).add(nodeList.get(i).val); 42 } 43 44 45 return resList; 46 } 47 48 }
LeetCode-Binary Tree Vertical Order Traversal
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原文地址：http://www.cnblogs.com/lishiblog/p/5725751.html
