Leet Code OJ 102. Binary Tree Level Order Traversal [Difficulty: Easy]

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题目:
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},
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return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

翻译:
给定一个二叉树,返回它的节点值的层序遍历(即从左到右,一层一层的)。

分析:
采用递归的方式,先返回根节点,然后递归调用左右节点,把左右节点的结果进行按照题目要求的形式拼装。另外注意一些递归过程中节点为空的情况。

代码:

/**
 * 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<List<Integer>> result=new ArrayList<>();
        if(root==null){
            return result;
        }
        List<Integer> one=new ArrayList<>();
        one.add(root.val);
        result.add(one);
        List<List<Integer>> left=levelOrder(root.left);
        List<List<Integer>> right=levelOrder(root.right);
        for(int i=0;i<left.size()||i<right.size();i++){
            List<Integer> item=new ArrayList<>();
            if(i<left.size()){
                for(Integer k:left.get(i)){
                    item.add(k);
                }
            }
            if(i<right.size()){
                for(Integer k:right.get(i)){
                    item.add(k);
                }
            }
            result.add(item);
        }
        return result;
    }
}











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