Binary Tree Zigzag Level Order Traversal——LeetCode
2015-04-21 21:47
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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
return its zigzag level order traversal as:
题目大意:给一个二叉树,将这个二叉树以zigzag形式输出,即一行以从左到右的顺序输出,下一行以从右到左的顺序输出……
解题思路:其实本质还是层次遍历,我们可以用cnt记录层数,奇数层从右到左,偶数层从左到右。用一个List记录这一层的数据,从左到右的顺序插入,从右到左的采用头插法插入即可。
Talk is cheap>>
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] ]
题目大意:给一个二叉树,将这个二叉树以zigzag形式输出,即一行以从左到右的顺序输出,下一行以从右到左的顺序输出……
解题思路:其实本质还是层次遍历,我们可以用cnt记录层数,奇数层从右到左,偶数层从左到右。用一个List记录这一层的数据,从左到右的顺序插入,从右到左的采用头插法插入即可。
Talk is cheap>>
public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) { return res; } int cnt = 0; Deque<TreeNode> queue = new ArrayDeque<>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> level = new ArrayList<>(); for (int i = 0; i < size; i++) { TreeNode node = queue.poll(); if ((cnt & 0x1) == 0) { level.add(node.val); } else { level.add(0, node.val); } if (node.left != null) { queue.offer(node.left); } if (node.right != null) { queue.offer(node.right); } } res.add(level); cnt++; } return res; }
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