LeetCode——Binary Tree Level Order Traversal II

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/bhlsheji/p/5305698.html

本文介绍如何通过层序遍历二叉树并将其结果反转来获取从叶子节点到根节点的层序遍历顺序。通过使用队列进行层序遍历，最后通过反转结果数组来实现目标。

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

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

return its bottom-up level order traversal as:

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

原题链接：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

题目：给定一个二叉树，从底到顶返回节点的层序遍历的值。（如，从左到右，一层一层）

思路：上一题的结果上面逆转一下就可以。全然不用出这题吧？

	public List<List<Integer>> levelOrderBottom(TreeNode root) {
		List<List<Integer>> list = new ArrayList<List<Integer>>();
		if (root == null)
			return list;
		Queue<TreeNode> queue = new LinkedList<TreeNode>();
		queue.add(root);
		while (!queue.isEmpty()) {
			List<Integer> li = new ArrayList<Integer>();
			int size = queue.size();
			for (int i = 0; i < size; i++) {
				TreeNode node = queue.poll();
				li.add(node.val);
				if (node.left != null)
					queue.add(node.left);
				if (node.right != null)
					queue.add(node.right);
			}
			list.add(li);
		}
		List<List<Integer>> ret = new ArrayList<List<Integer>>();
		for (int i = list.size() - 1; i >= 0; i--) {
			ret.add(list.get(i));
		}
		return ret;
	}

	// Definition for binary tree
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

转载于:https://www.cnblogs.com/bhlsheji/p/5305698.html