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

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;}