Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/
9 20
/
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
Solution:
1 /** 2 * Definition for binary tree 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>> zigzagLevelOrder(TreeNode root) { 12 List<List<Integer>> res = new ArrayList<List<Integer>>(); 13 if (root==null) return res; 14 15 List<List<TreeNode>> nodeSet = new ArrayList<List<TreeNode>>(); 16 List<TreeNode> oneLevel = new ArrayList<TreeNode>(); 17 nodeSet.add(oneLevel); 18 oneLevel.add(root); 19 boolean left = false; 20 int index = 0; 21 while (index<nodeSet.size()){ 22 List<TreeNode> curLevel = nodeSet.get(index); 23 oneLevel = new ArrayList<TreeNode>(); 24 for (int i=curLevel.size()-1;i>=0;i--){ 25 TreeNode curNode = curLevel.get(i); 26 if (left){ 27 if (curNode.left!=null) oneLevel.add(curNode.left); 28 if (curNode.right!=null)oneLevel.add(curNode.right); 29 } else { 30 if (curNode.right!=null) oneLevel.add(curNode.right); 31 if (curNode.left!=null) oneLevel.add(curNode.left); 32 } 33 } 34 if (oneLevel.size()>0) nodeSet.add(oneLevel); 35 left = !left; 36 index++; 37 } 38 39 for (int i=0;i<nodeSet.size();i++){ 40 oneLevel = nodeSet.get(i); 41 List<Integer> oneRes = new ArrayList<Integer>(); 42 for (int j=0;j<oneLevel.size();j++) 43 oneRes.add(oneLevel.get(j).val); 44 res.add(oneRes); 45 } 46 47 return res; 48 } 49 }
Solution 2:
We actually only need to store the nodes in the current level.
1 /** 2 * Definition for binary tree 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>> zigzagLevelOrder(TreeNode root) { 12 List<List<Integer>> resList = new ArrayList<List<Integer>>(); 13 if (root==null) return resList; 14 //root level, 15 List<Integer> oneLevel = new ArrayList<Integer>(); 16 List<TreeNode> nodeList = new ArrayList<TreeNode>(); 17 oneLevel.add(root.val); 18 resList.add(oneLevel); 19 nodeList.add(root); 20 boolean negOrder = true; 21 while (nodeList.size()>0){ 22 List<TreeNode> nextNode = new ArrayList<TreeNode>(); 23 //Visit the next level. 24 for (int i=nodeList.size()-1;i>=0;i--){ 25 TreeNode cur = nodeList.get(i); 26 if (negOrder){ 27 if (cur.right!=null) nextNode.add(cur.right); 28 if (cur.left!=null) nextNode.add(cur.left); 29 } else { 30 if (cur.left!=null) nextNode.add(cur.left); 31 if (cur.right!=null) nextNode.add(cur.right); 32 } 33 } 34 35 //If there are nodes in the next level, then put their value into results. 36 if (nextNode.size()>0){ 37 oneLevel = new ArrayList<Integer>(); 38 for (int i=0;i<nextNode.size();i++) oneLevel.add(nextNode.get(i).val); 39 resList.add(oneLevel); 40 } 41 42 nodeList = nextNode; 43 negOrder = !negOrder; 44 } 45 46 return resList; 47 48 } 49 }