Java [Leetcode 107]binary Tree level Order traversal II

Title Description:

Given a binary tree, return the bottom-up level order traversal of its nodes ' values. (ie, from the left-to-right, the level by level from the leaf to root).

For example:
Given binary Tree {3,9,20,#,#,15,7} ,

    3   /   9    /   7

Return its bottom-up level order traversal as:

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

Problem Solving Ideas:

Using the breadth-first search method, the queue method is used to traverse the leaf nodes of one layer at a time, and the leaves nodes of the next layer are added to the queue. Each time you traverse the end of a layer of nodes, the list of the layer nodes is placed before the list of the whole to achieve a bottom-to-high arrangement.

The code is as follows:

/** * 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>> levelorderbottom (TreeNode root) {    list< list<integer>> list = new linkedlist<list<integer>> ();    queue<treenode> queue = new linkedlist<treenode> ();    if (root = null)    return list;    Queue.offer (root);    while (!queue.isempty ()) {    int num = Queue.size ();    list<integer> levellist = new linkedlist<integer> ();    for (int i = 0; i < num; i++) {    if (Queue.peek (). Left! = null)    Queue.offer (Queue.peek (). left);    if (Queue.peek (). Right! = null)    Queue.offer (Queue.peek (). right);    Levellist.add (Queue.poll (). val);    }    List.add (0, levellist);    }    return list;    }}

　　

Java [Leetcode 107]binary Tree level Order traversal II