LeetCode 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}
,
DFS遍历方法:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 private: 12 vector<vector<int> > ret; 13 public: 14 void solve(int dep, TreeNode *root) 15 { 16 if (root == NULL) 17 return; 18 19 if (ret.size() > dep) 20 { 21 ret[dep].push_back(root->val); 22 } 23 else 24 { 25 vector<int> a; 26 a.push_back(root->val); 27 ret.push_back(a); 28 } 29 30 solve(dep + 1, root->left); 31 solve(dep + 1, root->right); 32 } 33 34 vector<vector<int> > levelOrder(TreeNode *root) { 35 // Start typing your C/C++ solution below 36 // DO NOT write int main() function 37 ret.clear(); 38 solve(0, root); 39 40 return ret; 41 } 42 };
BFS遍历打印
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 11 struct Node 12 { 13 TreeNode *node; 14 int level; 15 Node(){} 16 Node(TreeNode *n, int l):node(n), level(l){} 17 }; 18 19 class Solution { 20 private: 21 vector<vector<int> > ret; 22 public: 23 vector<vector<int> > levelOrder(TreeNode *root) { 24 // Start typing your C/C++ solution below 25 // DO NOT write int main() function 26 ret.clear(); 27 28 if (root == NULL) 29 return ret; 30 31 queue<Node> q; 32 33 q.push(Node(root, 0)); 34 35 vector<int> a; 36 int curLevel = -1; 37 38 while(!q.empty()) 39 { 40 Node node = q.front(); 41 if (node.node->left) 42 q.push(Node(node.node->left, node.level + 1)); 43 if (node.node->right) 44 q.push(Node(node.node->right, node.level + 1)); 45 46 if (curLevel != node.level) 47 { 48 if (curLevel != -1) 49 ret.push_back(a); 50 curLevel = node.level; 51 a.clear(); 52 a.push_back(node.node->val); 53 } 54 else 55 a.push_back(node.node->val); 56 57 q.pop(); 58 } 59 60 ret.push_back(a); 61 62 return ret; 63 } 64 };