103. Binary Tree Zigzag Level Order Traversal

Leetcode Tree Breath-first Search Stack

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]
]

分析

这道题目是LeetCode 102. Binary Tree Level Order Traversal的变形，其实增加一个判断即可，如果是从左往右，那么直接添加；如果是从右往左，那么先反转再添加。

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    List<List<Integer>> zigzag = new ArrayList<>();
    if (root == null) return zigzag;
    boolean isLeftToRight = true;
    Queue<TreeNode> queue = new LinkedList<>();
    queue.add(root);
    while(!queue.isEmpty()) {
        int size = queue.size();
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();
            list.add(node.val);
            if (node.left != null) queue.offer(node.left);
            if (node.right != null) queue.offer(node.right);
        }

        if (isLeftToRight){
            zigzag.add(list);
            isLeftToRight = false;
        } else {
            Collections.reverse(list);
            zigzag.add(list);
            isLeftToRight = true;
        }
    }

    return zigzag;
}