103. Binary Tree Zigzag Level Order Traversal

二刷。

BFS，基本習慣上用Iterative的做法來做，就是QUEUE。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<List<Integer>>();if(root == null) return res;Queue<TreeNode> q = new LinkedList<TreeNode>();q.add(root);boolean left = false;int cur = 1;int total = 0;TreeNode temp = null;List<Integer> tempList = new ArrayList<Integer>();while(!q.isEmpty()){temp = q.poll();cur--;tempList.add(temp.val);if(temp.left != null){q.add(temp.left);total++;}if(temp.right != null){q.add(temp.right);total++;}if(cur == 0){cur = total;total = 0;if(!left){res.add(new ArrayList<Integer>(tempList));left = !left;}else{Collections.reverse(tempList);res.add(new ArrayList<Integer>(tempList));left = !left;}tempList = new ArrayList<Integer>();}}return res;}
}

reverse一個LIST的函數是Collections.reverse(list)，不是list.reverse()。。那個是StringBuilder。

看一刷說還有Recursivley的做法。

然后試試發現還真行。。

public class Solution {public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<List<Integer>>();if(root == null) return res;helper(res,root,0);for(int i = 1; i < res.size();i+=2){Collections.reverse(res.get(i));}return res;}public void helper(List<List<Integer>> res, TreeNode root, int level){if(root == null) return;List<Integer> tempList = new ArrayList<Integer>();if(level > res.size()-1) res.add(new ArrayList<Integer>());res.get(level).add(root.val);helper(res,root.left,level+1);helper(res,root.right,level+1);}
}

beat 95%..

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三刷。

盡量加的時候就弄好順序，別最后排序。。

Recursion:

public class Solution {public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<>();if (root == null) return res;dfs(root, res, true, 0);return res;}public void dfs(TreeNode root, List<List<Integer>> res, boolean fromLeft, int level) {if (root == null) return;if (level == res.size()) {res.add(new ArrayList<>());}if (fromLeft) {res.get(level).add(root.val);} else {res.get(level).add(0,root.val);}dfs(root.left, res, !fromLeft, level + 1);dfs(root.right, res, !fromLeft, level + 1);}
}

Iteration:

public class Solution {public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<>();if (root == null) return res;res.add(new ArrayList<>());TreeNode temp = null;boolean fromLeft = true;int left = 1;       // how many nodes left in this levelint total = 0;int level = 0;Queue<TreeNode> q = new LinkedList<>();q.offer(root);while (!q.isEmpty()) {temp = q.poll();if (fromLeft) {res.get(level).add(temp.val);} else {res.get(level).add(0, temp.val);}if (temp.left != null) {q.offer(temp.left);total ++;}if (temp.right != null) {q.offer(temp.right);total ++;}if (-- left == 0) {left = total;total = 0;level ++;fromLeft = !fromLeft;if (!q.isEmpty()) {res.add(new ArrayList<>());}}}return res;}
}