LeetCode Binary Tree Level Order Traversal

传送门:https://oj.leetcode.com/problems/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).

题目大意:二叉树的层序遍历  

题目分析:题目要求按层次存入vector<vector<int>>中  这就要求在用队列遍历的同时, 还需要记录当前所在层数, 然后每层存一次。

我的解法:采用BFS, 使用队列, 并记录层数。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        int order = 0, level = 0, next = 0, count = 1;
        vector<vector<int>> sol;
        vector<int> row;
        queue<TreeNode*> q;
        TreeNode* b;
        if(root == NULL) return sol;
        q.push(root);
        while(!q.empty()){
            b = q.front();
            q.pop();
            row.push_back(b->val);
            order++;
            if(b->left != NULL){
                q.push(b->left);
                next++;
            }
            if(b->right != NULL){
                q.push(b->right);
                next++;
            }
            if(order == count){
                level++;
                order = 0;
                sol.push_back(row);
                count = next;
                next = 0;
                row.clear();
            }
        }
        return sol;
    }
};
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原文地址:https://www.cnblogs.com/ACFLOOD/p/4238547.html