39.2: Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2024-10-27 16:33:44 发布

转载最新推荐文章于 2024-10-27 16:33:44 发布 · 66 阅读

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原文链接：http://www.cnblogs.com/theonemars/p/4254207.html

本文介绍了一种二叉树的锯齿形层序遍历算法，通过改变每层节点的访问顺序实现从左到右，再从右到左的交替遍历。该算法使用了队列来辅助实现层序遍历，并通过布尔标志控制遍历方向，适用于二叉树的可视化展示和特定数据处理场景。

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/************************************************************************/
       /*       38:      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},

    3
   / \
9 20
    / \
   15   7

return its zigzag level order traversal as:

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

       * */

       /*** 树的按照层遍历，类似 BFS***********************************************************/

public List<List<Integer>> zigzagLevelOrder(TreeNode root) 
        {
             List<List<Integer>> result = new ArrayList<>();
                List<TreeNode> level = new ArrayList<>();
                level.add(root);
                boolean leftflag=true;
                while(true){
                    if (level.isEmpty() || level.get(0) == null){
                        break;
                    }
                    List<TreeNode> nextLevel = new ArrayList<>();
                    List<Integer> currentLevel = new ArrayList<>();
                    
                    if(leftflag) //left to right
                    {
                         for (TreeNode node : level){
                                currentLevel.add(node.val);
                                if (node.left != null) nextLevel.add(node.left);
                                if (node.right != null) nextLevel.add(node.right);
                            }
                    }
                    else
                    {
                        for(int i=level.size()-1;i>=0;i--)
                        {
                            currentLevel.add(level.get(i).val);
                            if (level.get(level.size()-1-i).left != null) nextLevel.add(level.get(level.size()-1-i).left);
                            if (level.get(level.size()-1-i).right != null) nextLevel.add(level.get(level.size()-1-i).right);
                        }
                    }
                    result.add(currentLevel);
                    level = nextLevel;
                    leftflag=!leftflag;
                }
                return result;
            
        }

转载于:https://www.cnblogs.com/theonemars/p/4254207.html