102. Binary Tree Level Order Traversal

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

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

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

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

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/** * 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>> levelOrder(TreeNode root) {                List<TreeNode> tree1 =new ArrayList <TreeNode> ();        List<List<Integer>> ret = new ArrayList <List<Integer>> ();        List<List<Integer>> ret2 = new ArrayList <List<Integer>> ();        if(root==null)return ret;        tree1.add(root);        while(tree1.size()>0){             List<TreeNode> tree2 =new ArrayList <TreeNode> ();            List <Integer> a = new ArrayList <Integer> ();            for(int i = 0;i<tree1.size();i++){                a.add(tree1.get(i).val);                if(tree1.get(i).left!=null)tree2.add(tree1.get(i).left);                if(tree1.get(i).right!=null)tree2.add(tree1.get(i).right);            }            tree1=tree2;            ret2.add(a);        }        // for(int i=ret2.size()-1;i>=0;i--){        //     ret.add(ret2.get(i));        // }        return ret2;        }}

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