Binary Tree Zigzag Level Order Traversal
描述
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]
]
分析
广度优先遍历，用一个bool记录是从左到右还是从右到左，每一层结束就翻转一下。
递归版
// Binary Tree Zigzag Level Order Traversal
// 递归版，时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List> zigzagLevelOrder(TreeNode root) {
List> result = new ArrayList<>();
traverse(root, 1, result, true);
return result;
}
private static void traverse(TreeNode root, int level, List> result,
boolean left_to_right) {
if (root == null) return;
if (level > result.size())
result.add(new ArrayList<>());
if (left_to_right)
result.get(level-1).add(root.val);
else
result.get(level-1).add(0, root.val);
traverse(root.left, level+1, result, !left_to_right);
traverse(root.right, level+1, result, !left_to_right);
}
}迭代版
// Binary Tree Zigzag Level Order Traversal
// 广度优先遍历，用一个bool记录是从左到右还是从右到左，每一层结束就翻转一下。
// 迭代版，时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List> zigzagLevelOrder(TreeNode root) {
List> result = new ArrayList<>();
Queue current = new LinkedList<>();
Queue next = new LinkedList<>();
boolean left_to_right = true;
if(root == null) {
return result;
} else {
current.offer(root);
}
while (!current.isEmpty()) {
ArrayList level = new ArrayList<>(); // elments in one level
while (!current.isEmpty()) {
TreeNode node = current.poll();
level.add(node.val);
if (node.left != null) next.offer(node.left);
if (node.right != null) next.offer(node.right);
}
if (!left_to_right) Collections.reverse(level);
result.add(level);
left_to_right = !left_to_right;
// swap
Queue tmp = current;
current = next;
next = tmp;
}
return result;
}
}