leetcode-Binary Tree Level Order Traversal 二叉树层序遍历

leetcode-Binary Tree Level Order Traversal 二叉树层序遍历

#include<stdio.h>#include<queue>using namespace std;typedef struct BiTree{int val;struct BiTree *lchild;struct BiTree *rchild;}BiTree;void main(){BiTree *root;queue<BiTree *> q;BiTree *t;q.push(root);int size;//核心代码while(!q.empty()){size=q.size();while(size--){t=q.front();q.pop();printf("%d ",t->val);if(t->lchild!=NULL)q.push(t->lchild);if(t->rchild!=NULL)q.push(t->rchild);}}}

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).

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

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/** * 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) {                                vector<int>level;        vector<vector<int>>res;        int size;        TreeNode *t;                if(root==NULL)            return res;                queue<TreeNode*> q;        q.push(root);                while(!q.empty())        {            size=q.size();                        while(size--)            {                t=q.front();                q.pop();                                level.push_back(t->val);                                if(t->left!=NULL)                    q.push(t->left);                                if(t->right!=NULL)                    q.push(t->right);            }                        res.push_back(level);            level.clear();                    }        return res;    }};