leetcode102. Binary Tree Level Order Traversal-优快云博客

本文介绍了二叉树层次遍历的两种实现方法：BFS（广度优先搜索）和DFS（深度优先搜索）。通过队列和递归的方式分别实现了从上到下、从左到右的层次遍历。

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102. Binary Tree Level Order Traversal

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]
]

解法一

BFS: queue进队列offer，出队列poll，将数的每一层先进入队列，再将队列中的数依次执行运算，是否有左右孩子，都加入到队列中。直到队列为空。

/**
 * 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>> ret = new ArrayList<List<Integer>>();
        if (root == null) {
            return ret;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> level = new ArrayList<Integer>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode head = queue.poll();
                level.add(head.val);
                if (head.left != null) {
                    queue.offer(head.left);
                }
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            ret.add(level);
        }

        return ret;
    }
}

这里写图片描述

解法二

DFS

/**
 * 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>> ret = new ArrayList<List<Integer>>();
        levelAdd(root, ret, 0);
        return ret;
    }
    private void levelAdd(TreeNode root, List<List<Integer>> ret, int level) {
        if (root == null) {
            return;
        }

        if (ret.size() <= level) {
            ret.add(new ArrayList<Integer>());
        }
        ret.get(level).add(root.val);
        levelAdd(root.left, ret, level + 1);
        levelAdd(root.right, ret, level + 1);
    }
}

这里写图片描述