103. Binary Tree Zigzag Level Order Traversal

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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 [3,9,20,null,null,15,7],

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2
3
4
5
  3
 / \
9  20
  /  \
 15   7

return its zigzag level order traversal as:

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2
3
4
5
[
  [3],
  [20,9],
  [15,7]
]

解法1：BFS

BFS 然后reverse间隔行就可以了。
C++

1

Java

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/**
 * 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> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        int level = 1;
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode current = queue.poll();
                temp.add(current.val);
                if (current.left != null) {
                    queue.offer(current.left);
                }
                if (current.right != null) {
                    queue.offer(current.right);
                }
            }
            if (level % 2 == 0) {
                // reverse the temp
                for (int i = 0, j = temp.size() - 1; i < j; i++, j--) {
                    int buffer = temp.get(i);
                    temp.set(i, temp.get(j));
                    temp.set(j, buffer);
                }
            }
            res.add(temp);
            level++;
        }
        return res;
    }
}