Convert Binary Tree to Linked Lists by Depth

最新推荐文章于 2024-09-03 15:28:22 发布

转载最新推荐文章于 2024-09-03 15:28:22 发布 · 141 阅读

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原文链接：http://www.cnblogs.com/codingEskimo/p/6999502.html

本文介绍了一种将二叉树按层序遍历并转换为链表的方法。通过创建哑节点和链表节点进行遍历，最终得到每层节点构成的链表。适用于需要对二叉树进行层序遍历并转换成链表结构的场景。

This question is the same as Binary Tree Treverse in Level Order. Instead of a list, we need ListNode to remeber the result. The tricky part is that we need to create a dummy node and listNode for traval. Keep moveing the listNode to add next and then add dummy.next to the result list

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
     * @param root the root of binary tree
     * @return a lists of linked list
     */
    public List<ListNode> binaryTreeToLists(TreeNode root) {
        // Write your code here
        List<ListNode> rst = new ArrayList<ListNode>();
        if (root == null) {
            return rst;
        }
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        while (!q.isEmpty()) {
            ListNode dummy = new ListNode(0);
            ListNode listNode = dummy;
            int n = q.size();
            for (int i = 0; i < n; i++) {
                TreeNode node = q.poll();
                listNode.next = new ListNode(node.val); 
                listNode = listNode.next;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            rst.add(dummy.next);
        }
        return rst;
    }
}

转载于:https://www.cnblogs.com/codingEskimo/p/6999502.html