LeetCode OJ:Binary Tree Level Order Traversal
2014-01-22 17:43
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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
{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.
算法思想:层序遍历
/** * 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<vector<int>> result; if(!root)return result; vector<int> t; int curLev=1; int nextLev=0; queue<TreeNode *> que; que.push(root); while(!que.empty()){ TreeNode *cur=que.front(); que.pop(); t.push_back(cur->val); if(cur->left){ nextLev++; que.push(cur->left); } if(cur->right){ nextLev++; que.push(cur->right); } if(--curLev==0){ result.push_back(t); t.clear(); curLev=nextLev; nextLev=0; } } return result; } };
递归版
/** * 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<vector<int>> result; traverse(root,1,result); return result; } void traverse(TreeNode *root,size_t level,vector<vector<int>> &result){ if(!root)return; if(level>result.size()) result.push_back(vector<int>()); result[level-1].push_back(root->val); traverse(root->left,level+1,result); traverse(root->right,level+1,result); } };
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