leetcode 107. Binary Tree Level Order Traversal II(BFS)(Java和C++)

最新推荐文章于 2024-02-12 01:19:39 发布

原创最新推荐文章于 2024-02-12 01:19:39 发布 · 696 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #bfs

JAVA 同时被 3 个专栏收录

68 篇文章

订阅专栏

leetcode

47 篇文章

订阅专栏

C++

9 篇文章

订阅专栏

本文介绍了一种从叶子节点到根节点的层次遍历二叉树的方法，并提供了Java和C++实现。通过广度优先搜索策略，可以有效地获取树的每一层节点值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述：

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).

思路：

利用广度优先搜索

Java代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public List<List<Integer>> levelOrderBottom(TreeNode root){
        List<List<Integer>> result = new ArrayList<>();
        if(root == null) return result;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while(!queue.isEmpty()){
            List<Integer> list = new ArrayList<>();
            int num = queue.size();
            for(int i = 0; i < num; i++){
                TreeNode n = queue.poll();
                if(n.left != null) queue.add(n.left);
                if(n.right != null) queue.add(n.right);
                list.add(n.val);
            }
            result.add(list);
        }       
        Collections.reverse(result);
        return result;      
}

public static void main(String[] args) {
         RomanToInt a=new RomanToInt();  
         //int[] at = {1,2,2,3,1};
         String at = "abcdefg";
         TreeNode root = new TreeNode(3);
         root.left = new TreeNode(9);
         root.right = new TreeNode(20);
         root.right.left = new TreeNode(15);
         root.right.right = new TreeNode(7);
         System.out.println(a.levelOrderBottom(root));          
    }

C++代码：

#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;


struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) :val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        if (root == NULL) return result;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty())
        {
            int n = q.size();
            vector<int> v;
            for (int i = 0; i < n; i++) {               
                TreeNode* n = q.front();
                q.pop();
                if (n->left != NULL) q.push(n->left);
                if (n->right != NULL) q.push(n->right);
                v.push_back(n->val);
            }
            result.push_back(v);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};


int main()
{
    TreeNode *root = new TreeNode(3);
    root->left = new TreeNode(9);
    root->right = new TreeNode(20);
    root->right->left = new TreeNode(15);
    root->right->right = new TreeNode(7);
    Solution a;
    vector<vector<int>> result = a.levelOrderBottom(root);
    for (int i = 0; i < result.size(); i++) {
        for (int j = 0; j < result[i].size(); j++) {
            cout << result[i][j]<<" ";
        }
        cout << endl;
    }
    system("pause");
    return 0;
}