【LeetCode】102. Binary Tree Level Order Traversal (2 solutions)
2014-11-30 20:12
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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
return its level order traversal as:
解法一:递归
参考了Discussion中stellari的做法,递归进行层次遍历,并将每个level对应于相应的vector。
![](http://images.cnitblog.com/blog2015/458814/201504/042101256072531.jpg)
解法二:
层次遍历,层数使用level来记录。同层装入同一个vector。
当进入新的一层时,将上层的vector保存,并清空。
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] ]
解法一:递归
参考了Discussion中stellari的做法,递归进行层次遍历,并将每个level对应于相应的vector。
/** * 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> > result; void levelTra(TreeNode *root, int level) { if(root == NULL) return; if(level == result.size()) { vector<int> v; result.push_back(v); } result[level].push_back(root->val); levelTra(root->left, level+1); levelTra(root->right, level+1); } vector<vector<int> > levelOrder(TreeNode *root) { levelTra(root, 0); return result; } };
![](http://images.cnitblog.com/blog2015/458814/201504/042101256072531.jpg)
解法二:
层次遍历,层数使用level来记录。同层装入同一个vector。
当进入新的一层时,将上层的vector保存,并清空。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ struct Node { TreeNode* tNode; int level; Node(TreeNode* newtNode, int newlevel): tNode(newtNode), level(newlevel) {} }; class Solution { public: vector<vector<int> > levelOrder(TreeNode *root) { vector<vector<int> > ret; if(!root) return ret; // push root Node* rootNode = new Node(root, 0); queue<Node*> Nqueue; Nqueue.push(rootNode); vector<int> cur; int curlevel = 0; while(!Nqueue.empty()) { Node* frontNode = Nqueue.front(); Nqueue.pop(); if(frontNode->level > curlevel) { ret.push_back(cur); cur.clear(); curlevel = frontNode->level; } cur.push_back(frontNode->tNode->val); if(frontNode->tNode->left) { Node* leftNode = new Node(frontNode->tNode->left, frontNode->level+1); Nqueue.push(leftNode); } if(frontNode->tNode->right) { Node* rightNode = new Node(frontNode->tNode->right, frontNode->level+1); Nqueue.push(rightNode); } } ret.push_back(cur); return ret; } };
![](http://images.cnitblog.com/blog2015/458814/201504/041132344513459.jpg)
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