Binary Tree Zigzag Level Order Traversal
update Aug 31, 2017 15:54
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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
Basic Idea:
Level order traverse,但是按照 “left-right”,“right-left” ,... 的顺序进行。解法非常简单,只要正常按照从左到右的方向进行,另外维持一个boolean标识当前的方向,如果是右向左,则把当前 level 的 vals 反向之后再加入 res;
Java Code:
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
Deque<TreeNode> queue = new LinkedList<>();
boolean leftToRight = true;
queue.addFirst(root);
while (! queue.isEmpty()) {
int size = queue.size();
List<Integer> level = new ArrayList<>();
for (int i = 0; i < size; ++i) {
TreeNode node = queue.removeLast();
level.add(node.val);
if (node.left != null) queue.addFirst(node.left);
if (node.right != null) queue.addFirst(node.right);
}
if (! leftToRight) Collections.reverse(level);
res.add(level);
leftToRight ^= true;
}
return res;
}
}