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  • 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;
        }
    };
    View Code
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    12-30
    2016-12-29
  • 原文地址:https://www.cnblogs.com/ACFLOOD/p/4238547.html
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