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] ]
[解题思路]
本题是二叉树层序遍历的变形，二叉树的层序遍历通过使用queue来实现，
一开始我也试图用queue来解这题，第二层很好解决，但到第三层时无法实现从左到右遍历该层
上网搜索了下，发现该题可以使用栈来解决，通过分析执行结果确实是后进先出
这里通过定义leftToRight来表示是从左到右还是从右到左
从左到右：先加left后加right
从右到左：先加right后加left
1 public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) { 2 // Start typing your Java solution below 3 // DO NOT write main() function 4 ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); 5 if(root == null){ 6 return result; 7 } 8 Stack<TreeNode> curLvl = new Stack<TreeNode>(); 9 Stack<TreeNode> nextLvl = new Stack<TreeNode>(); 10 boolean leftToRight = true; 11 curLvl.push(root); 12 ArrayList<Integer> output = new ArrayList<Integer>(); 13 while(!curLvl.empty()){ 14 TreeNode curNode = curLvl.pop(); 15 output.add(curNode.val); 16 if(leftToRight){ 17 if(curNode.left != null){ 18 nextLvl.add(curNode.left); 19 } 20 if(curNode.right != null){ 21 nextLvl.add(curNode.right); 22 } 23 } else{ 24 if(curNode.right != null){ 25 nextLvl.add(curNode.right); 26 } 27 if(curNode.left != null){ 28 nextLvl.add(curNode.left); 29 } 30 } 31 if(curLvl.empty()){ 32 result.add(output); 33 output = new ArrayList<Integer>(); 34 leftToRight = !leftToRight; 35 curLvl.addAll(nextLvl); 36 nextLvl.clear(); 37 } 38 } 39 return result; 40 }