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] ]
实现代码:
class Solution { public: vector<vector<int> > levelOrder(TreeNode *root) { vector<vector<int> > res; if(root==NULL) return res; queue<TreeNode *> temp; temp.push(root); while(!temp.empty()) { vector<int> curlevel; queue<TreeNode *> nextlevel; while(!temp.empty()) { TreeNode* node = temp.front(); temp.pop(); curlevel.push_back(node->val); if (node->left != NULL) nextlevel.push(node->left); if (node->right != NULL) nextlevel.push(node->right); } res.push_back(curlevel); temp = nextlevel; } return res; } };
Given a binary tree, return the bottom-up level order traversal of its nodes‘ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7}
,
3 / 9 20 / 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
实现代码:
class Solution { public: vector<vector<int> > levelOrderBottom(TreeNode *root) { vector<vector<int> > res; if(root==NULL) return res; queue<TreeNode *> temp; temp.push(root); while(!temp.empty()) { vector<int> curlevel; queue<TreeNode *> nextlevel; while(!temp.empty()) { TreeNode* node = temp.front(); temp.pop(); curlevel.push_back(node->val); if (node->left != NULL) nextlevel.push(node->left); if (node->right != NULL) nextlevel.push(node->right); } res.insert(res.begin(),curlevel); temp = nextlevel; } return res; } };
原文地址:http://blog.csdn.net/wolongdede/article/details/44778341