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] ]
先输出成arraylist，然后再反转即可。
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 ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) { 12 // IMPORTANT: Please reset any member data you declared, as 13 // the same Solution instance will be reused for each test case. 14 ArrayList<TreeNode> queue = new ArrayList<TreeNode>(); 15 ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); 16 if(root == null) return result; 17 queue.add(root); 18 TreeNode nl = new TreeNode(Integer.MIN_VALUE); 19 queue.add(nl); 20 ArrayList<Integer> cur = new ArrayList<Integer>(); 21 while(queue.size() != 1){ 22 TreeNode rn = queue.remove(0); 23 if(rn.val == Integer.MIN_VALUE){ 24 queue.add(nl); 25 result.add(cur); 26 cur = new ArrayList<Integer>(); 27 } 28 else{ 29 if(rn.left != null) queue.add(rn.left); 30 if(rn.right != null) queue.add(rn.right); 31 cur.add(rn.val); 32 } 33 } 34 result.add(cur); 35 for (int i=0; i<result.size(); i++) 36 if (i % 2 == 1) reverse(result,i); 37 return result; 38 } 39 public void reverse(ArrayList<ArrayList<Integer>> result, int num){ 40 int len = result.get(num).size(); 41 for(int i = 0; i < len / 2; i ++){ 42 int tmp = result.get(num).get(i); 43 result.get(num).set(i, result.get(num).get(len - 1 - i)); 44 result.get(num).set(len - 1 - i, tmp); 45 } 46 } 47 }
第三遍：
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 ArrayList<List<Integer>> result = new ArrayList<List<Integer>>(); 13 if(root == null) return result; 14 LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); 15 queue.add(root); 16 TreeNode nil = new TreeNode(Integer.MAX_VALUE); 17 queue.add(nil); 18 boolean even = false; 19 ArrayList<Integer> row = new ArrayList<Integer>(); 20 while(queue.size() != 1){ 21 TreeNode tmp = queue.poll(); 22 if(tmp.val != Integer.MAX_VALUE){ 23 row.add(tmp.val); 24 if(tmp.left != null) queue.add(tmp.left); 25 if(tmp.right != null) queue.add(tmp.right); 26 } else { 27 if(even) reverseRow(row); 28 result.add(row); 29 row = new ArrayList<Integer>(); 30 queue.add(nil); 31 even = !even; 32 } 33 } 34 if(even) reverseRow(row); 35 result.add(row); 36 return result; 37 } 38 39 public void reverseRow(ArrayList<Integer> row){ 40 int len = row.size(); 41 for(int i = 0; i < len / 2; i ++){ 42 int tmp = row.get(i); 43 row.set(i, row.get(len - 1 - i)); 44 row.set(len - 1 - i, tmp); 45 } 46 } 47 }