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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] ]
难度95，调试老是TLE，看了一下别人的做法，没懂为啥我的超时了。参考这道题其实还是树的层序遍历Binary Tree Level Order Traversal，如果不熟悉的朋友可以先看看哈。不过这里稍微做了一点变体，就是在遍历的时候偶数层自左向右，而奇数层自右向左。在Binary Tree Level Order Traversal中我们是维护了一个队列来完成遍历，而在这里为了使每次都倒序出来，我们很容易想到用栈的结构来完成这个操作。有一个区别是这里我们需要一层一层的来处理（原来可以按队列插入就可以，因为后进来的元素不会先处理），所以会同时维护新旧两个栈，一个来读取，一个存储下一层结点。总体来说还是一次遍历完成，所以时间复杂度是O(n)，空间复杂度最坏是两层的结点，所以数量级还是O(n)（满二叉树最后一层的结点是n/2个）。代码如下：
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 ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>>(); 13 if (root == null) return lists; 14 LinkedList<TreeNode> stack = new LinkedList<TreeNode>(); 15 ArrayList<Integer> list = new ArrayList<Integer>(); 16 stack.push(root); 17 list.add(root.val); 18 lists.add(list); 19 int level = 1; 20 while (!stack.isEmpty()) { 21 list = new ArrayList<Integer>(); 22 LinkedList<TreeNode> newstack = new LinkedList<TreeNode>(); 23 while (!stack.isEmpty()) { 24 TreeNode cur = stack.pop(); 25 if (level % 2 == 0) { 26 if (cur.left != null) { 27 list.add(cur.left.val); 28 newstack.push(cur.left); 29 } 30 if (cur.right != null) { 31 list.add(cur.right.val); 32 newstack.push(cur.right); 33 } 34 } 35 else { 36 if (cur.right != null) { 37 list.add(cur.right.val); 38 newstack.push(cur.right); 39 } 40 if (cur.left != null) { 41 list.add(cur.left.val); 42 newstack.push(cur.left); 43 } 44 } 45 } 46 level++; 47 if (list.size()>0) { 48 lists.add(list); 49 } 50 stack = newstack; 51 } 52 return lists; 53 } 54 }
贴一下我的做法，不知为何TLE
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 ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>>(); 13 if (root == null) return lists; 14 LinkedList<TreeNode> stack = new LinkedList<TreeNode>(); 15 LinkedList<TreeNode> newstack = new LinkedList<TreeNode>(); 16 stack.add(root); 17 int PNuminQ = 1; 18 int CNuminQ = 0; 19 int level = 0; 20 ArrayList<Integer> list = new ArrayList<Integer>(); 21 while (!stack.isEmpty()) { 22 TreeNode cur = stack.pop(); 23 PNuminQ--; 24 list.add(cur.val); 25 if (level%2 == 0) { 26 if (root.left != null) { 27 newstack.push(root.left); 28 CNuminQ++; 29 } 30 if (root.right != null) { 31 newstack.push(root.right); 32 CNuminQ++; 33 } 34 } 35 else if (level%2 == 1) { 36 if (root.right != null) { 37 newstack.push(root.right); 38 CNuminQ++; 39 } 40 if (root.left != null) { 41 newstack.push(root.left); 42 CNuminQ++; 43 } 44 } 45 46 if (PNuminQ == 0) { 47 PNuminQ = CNuminQ; 48 CNuminQ = 0; 49 level++; 50 lists.add(list); 51 list = new ArrayList<Integer>(); 52 stack = newstack; 53 newstack = new LinkedList<TreeNode>(); 54 } 55 } 56 return lists; 57 } 58 }
Leetcode Binary Tree Zigzag level Order Traversal
标签：des style blog http color io ar for div
原文地址：http://www.cnblogs.com/EdwardLiu/p/3974818.html
