【LeetCode】 102. Binary Tree Level Order Traversal 二叉树的层序遍历(Medium)(JAVA)

【LeetCode】 102. Binary Tree Level Order Traversal 二叉树的层序遍历(Medium)(JAVA)

题目地址: https://leetcode.com/problems/binary-tree-level-order-traversal/

题目描述:

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

题目大意

给你一个二叉树,请你返回其按 层序遍历 得到的节点值。 (即逐层地,从左到右访问所有节点)。

解题方法

用一个列表记录当前层级的 TreeNode,然后记录总共有多少个

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        List<Integer> cur = new ArrayList<>();
        List<TreeNode> nodes = new ArrayList<>();
        nodes.add(root);
        int count = 1;
        while (nodes.size() > 0) {
            TreeNode temp = nodes.remove(0);
            cur.add(temp.val);
            if (temp.left != null) nodes.add(temp.left);
            if (temp.right != null) nodes.add(temp.right);
            count--;
            if (count == 0) {
                res.add(new ArrayList<>(cur));
                cur = new ArrayList<>();
                count = nodes.size();
            } 
        }
        return res;
    }
}

执行用时 : 2 ms, 在所有 Java 提交中击败了 23.91% 的用户
内存消耗 : 40.3 MB, 在所有 Java 提交中击败了 5.71% 的用户

猜你喜欢

转载自blog.csdn.net/qq_16927853/article/details/105775142