Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2024-01-19 11:15:23 发布

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本文介绍了一种zigzag级序遍历二叉树的方法，通过使用inOrder遍历并存储结果到ArrayList中，然后根据层级反转奇数行来实现zigzag遍历效果。

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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

思路：用inOrder 来遍历这个树，把结果存在ArrayList里面，最后把奇数行的反转。
易错点：在ret.size() < cur + 1 时，创建的新list 要添加进 ret里面。还有反转时候是 size - 1 - j;

/**
 * Definition for binary tree
 * 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>> ret = new ArrayList<List<Integer>>();
        inOrder(root, 0, ret);
        for(int i = 0; i < ret.size(); i++){
            if(i % 2 == 1){
                List<Integer> level = ret.get(i);
                for(int j = 0; j < level.size() /2; j++){
                    swap(level, j, level.size() - 1 - j);
                }
            }
        }
        return ret;
    }
    
    private void swap(List<Integer> level, int a, int b){
        int temp = level.get(a);
        level.set(a, level.get(b));
        level.set(b, temp);
    }
    
    private void inOrder(TreeNode root, int cur, List<List<Integer>> ret){
        if(root == null)
            return;
        List<Integer> level;
        if(ret.size() < cur + 1){
            level = new ArrayList<Integer>();
            ret.add(level);
        }else{
            level = ret.get(cur);
        }
        level.add(root.val);
        inOrder(root.left, cur + 1, ret);
        inOrder(root.right, cur + 1, ret);
    }
}