【LeetCode从零单排】No102 Binary Tree Level Order Traversal

Given a binary tree, return thelevel ordertraversal 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]]

基本二叉树这块用的都是递归，，现在还是有点蒙。

代码

/** * 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>> levelOrder(TreeNode root) {List<List<Integer>> al=new ArrayList<List<Integer>>();levelHelper(al,root,0);return al;}public void levelHelper(List<List<Integer>> res,TreeNode root,int height){if(root==null) return;if(height>=res.size()){res.add(new ArrayList<Integer>());}res.get(height).add(root.val);levelHelper(res,root.left,height+1);levelHelper(res,root.right,height+1); }}

代码下载：https://github.com/jimenbian/GarvinLeetCode

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