[LeetCode]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]
]

简单的层次遍历,两个栈


/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
    		vector<vector<int> > ret;
			ret.clear();
			if(root == NULL)
				return ret;
			stack<TreeNode*> S;
			stack<TreeNode*> T;
			S.push(root);
			while(!S.empty()){
				vector<int> tmp;
				while(!S.empty()){
					TreeNode* p = S.top();
					S.pop();
					tmp.push_back(p->val);
					if(p->left) 
						T.push(p->left);
					if(p->right)
						T.push(p->right);
				}
				ret.push_back(tmp);
				if(T.empty())
					return ret;
				while(!T.empty())
				{
					TreeNode *p = T.top();
					T.pop();
					S.push(p);
				}
			}
    }
};


又写了一个队列的,用NULL指针把层次分隔开。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        	vector<vector<int> > ret;
			ret.clear();
			if(root == NULL)
				return ret;
			queue<TreeNode*> S;
			S.push(root);
			S.push(NULL);
			vector<int> tmp;
			while(!S.empty()){
				//travesal current level
				TreeNode* p = S.front();
				S.pop();
				if(p!=NULL)
				{
					tmp.push_back(p->val);
					if(p->left) 
						S.push(p->left);
					if(p->right)
						S.push(p->right);
				}else{
					if(!tmp.empty())
					{	
						S.push(NULL);
						ret.push_back(tmp);
						tmp.clear();
					}			  			
				}
			}
			return ret;
    }
};


posted @ 2012-11-06 15:30  程序员杰诺斯  阅读(114)  评论(0编辑  收藏  举报