LeetCode 103. Binary Tree Zigzag Level Order Traversal（二叉树之字形遍历）

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本文介绍如何解决LeetCode上的103题，即二叉树的锯齿形层次遍历。通过广度优先搜索(BFS)策略，实现节点值的交替左右遍历，从而达到题目要求的效果。

原题网址：https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/

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},

return its zigzag level order traversal as:

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

方法：广度优先搜索。

/**
 * 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>> results = new ArrayList<>();
        if (root == null) return results;
        List<TreeNode> level = new ArrayList<>(1);
        level.add(root);
        boolean forward = true;
        while (!level.isEmpty()) {
            List<Integer> result = new ArrayList<>();
            results.add(result);
            for(int i=0; i<level.size(); i++) result.add(level.get(i).val);
            
            List<TreeNode> next = new ArrayList<>();
            for(int i=level.size()-1; i>=0; i--) {
                if (forward) {
                    if (level.get(i).right != null) next.add(level.get(i).right);
                    if (level.get(i).left != null) next.add(level.get(i).left);
                } else {
                    if (level.get(i).left != null) next.add(level.get(i).left);
                    if (level.get(i).right != null) next.add(level.get(i).right);
                }
            }
            level = next;
            forward = !forward;
        }
        return results;
    }
}