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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] ]
层次遍历， 比较难处理的就是怎么判断是每层的队尾
有这么几个方法：
1. 用一个 current point来指示当前层，然后每次压入的时候不断更新，用一个tail来标记父节点的last pointer，当父节点走到最后了，把current赋给tail
1 class Solution { 2 public: 3 vector<vector<int> > levelOrder(TreeNode* root) { 4 vector<vector<int> > v; 5 vector<int> level; 6 if (root == NULL) 7 return v; 8 if (root ->left == NULL && root ->right == NULL) 9 { 10 level.push_back(root->val); 11 v.push_back(level); 12 return v; 13 } 14 queue<TreeNode*> q; 15 q.push(root); 16 TreeNode *trail,*cur; // trail record the last item in each level 17 trail = cur = root; 18 19 while(!q.empty()) 20 { 21 TreeNode* pointer = q.front(); 22 q.pop(); 23 level.push_back(pointer->val); 24 if (pointer->left != NULL) 25 { 26 cur = pointer->left; 27 q.push(pointer->left); 28 } 29 if (pointer->right != NULL) 30 { 31 cur = pointer->right; 32 q.push(pointer->right); 33 } 34 if (pointer == trail) 35 { 36 trail = cur; 37 v.push_back(level); 38 level.clear(); 39 } 40 } 41 return v; 42 } 43 };
2）第二个方法是，用NULL做一个，把NULL也压入queue里面，到访问到NULL的时候，就意味着是这层的最后一个
1 class Solution { 2 public: 3 vector<vector<int> > levelOrder(TreeNode *root) { 4 vector<vector<int> > result; 5 if (!root) return result; 6 queue<TreeNode*> q; 7 q.push(root); 8 q.push(NULL); 9 vector<int> cur_vec; 10 while(!q.empty()) { 11 TreeNode* t = q.front(); 12 q.pop(); 13 if (t==NULL) { 14 result.push_back(cur_vec); 15 cur_vec.resize(0); 16 if (q.size() > 0) { 17 q.push(NULL); 18 } 19 } else { 20 cur_vec.push_back(t->val); 21 if (t->left) q.push(t->left); 22 if (t->right) q.push(t->right); 23 } 24 } 25 return result; 26 } 27 };
3) preorder 用递归做，判断何时新加一个vector
1 vector<vector<int>> ret; 2 3 void buildVector(TreeNode *root, int depth) 4 { 5 if(root == NULL) return; 6 if(ret.size() == depth) 7 ret.push_back(vector<int>()); 8 9 ret[depth].push_back(root->val); 10 buildVector(root->left, depth + 1); 11 buildVector(root->right, depth + 1); 12 } 13 14 vector<vector<int> > levelOrder(TreeNode *root) { 15 buildVector(root, 0); 16 return ret; 17 }
4) 可以在压入队列的时候同时压入层次的数字
Leetcode 102 Binary Tree Level Order Traversal
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原文地址：http://www.cnblogs.com/zhuguanyu33/p/4617590.html
