leetcode-Binary Tree Level Order Traversal 二叉樹層序遍歷

leetcode-Binary Tree Level Order Traversal 二叉樹層序遍歷

#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;


	//核心代碼
	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;
    }
};