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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] ]
BFS思路：层序遍历用队列的先进先出来实现。最开始的想法是想在一个队列中通过统计某层得节点个数来完成把每一层存储到vector中，发现搞不出来。后来看到有人的代码用两个队列来实现~确实是个好办法。
pre队列装上一层得节点，然后把pre中节点的左右孩子全部放到cur中，放一次，就把pre中节点存储在temp中，然后从pre中pop，当pre为空时，那么就把temp放到结果res中，然后清空temp准备存储下一层，最后把cur赋给pre，清空cur。进而重复上面的步骤。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void qClear(queue<TreeNode*>& a) { while (!a.empty()) { a.pop(); } } vector<vector<int>> levelOrder(TreeNode *root) { vector<vector<int>> res; vector<int> temp; if(root==NULL) return res; queue<TreeNode*> pre,cur; pre.push(root); do { TreeNode* node=pre.front(); if(node->left!=NULL){ cur.push(node->left); } if(node->right!=NULL){ cur.push(node->right); } temp.push_back(pre.front()->val); pre.pop(); if(pre.empty()){//pre中所有节点清空 res.push_back(temp); temp.clear(); pre=cur; qClear(cur); } }while (!pre.empty()); return res; } };
DFS（转载，原文链接）：
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { private: vector<vector<int> > ret; public: void solve(int dep, TreeNode *root) { if (root == NULL) return; if (ret.size() > dep) { ret[dep].push_back(root->val); } else { vector<int> a; a.push_back(root->val); ret.push_back(a); } solve(dep + 1, root->left); solve(dep + 1, root->right); } vector<vector<int> > levelOrder(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function ret.clear(); solve(0, root); return ret; } };
Binary Tree Level Order Traversal
标签：des style blog http io color ar os for
原文地址：http://www.cnblogs.com/fightformylife/p/4078200.html
