Leetcode bfs&dfs Binary Tree Postorder Traversal

Binary Tree Level Order Traversal

Total Accepted: 20571 Total
Submissions: 66679My Submissions

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.

题意：给定一棵二叉树，返回按层遍历的结果

思路1：bfs，定义一个新的struct，记录指针向节点的指针和每个节点所在的层

思路2：dfs

递归函数：

void levelOrder(TreeNode *root, int level, vector<vector<int> >&result)

表示把根为root的树按层存放在result中，其中level表示当前的层数

复杂度：

bfs 时间O(n) ，空间O(n)

dfs 时间O(n) ，空间O(log n)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    struct NodeWithLevel{
    	TreeNode *p;
    	int level;
    	NodeWithLevel(TreeNode *pp, int l):p(pp), level(l){}
    };

    //思路1
     vector<vector<int> > levelOrder(TreeNode *root){
    	vector<vector<int> > result;
    	queue<NodeWithLevel > q;
    	q.push(NodeWithLevel(root, 0));
    	while(!q.empty()){
    		NodeWithLevel cur = q.front();q.pop();
    		TreeNode *t = cur.p;
    		if(t != NULL){

    			if(result.size() <= cur.level) {
    				vector<int> v;
    				v.push_back(t->val);
    				result.push_back(v);
    			}else result[cur.level].push_back(t->val);

    			q.push(NodeWithLevel(t->left, cur.level + 1));
    			q.push(NodeWithLevel(t->right, cur.level + 1));
    		}
    	}
    	return result;
    }
};

class Solution {
public:
    struct NodeWithLevel{
    	TreeNode *p;
    	int level;
    	NodeWithLevel(TreeNode *pp, int l):p(pp), level(l){}
    };

    //思路2
    void levelOrder(TreeNode *root, int level, vector<vector<int> >&result)
    {
    	if(!root) return ;
    	if(result.size() <= level) {
    		vector<int> v;
    		v.push_back(root->val);
    		result.push_back(v);
    	}else result[level].push_back(root->val);

    	levelOrder(root->left, level + 1, result);
    	levelOrder(root->right, level + 1, result);
    }

    vector<vector<int> > levelOrder(TreeNode *root){
    	vector<vector<int> >result;
    	levelOrder(root, 0, result);
    	return result;
    }
};