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Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
不算难，使用stack即可。
1 /** 2 * Definition for a binary tree node. 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 public: 12 vector<int> preorderTraversal(TreeNode* root) { 13 vector<int> result; 14 TreeNode* p; 15 stack<TreeNode*> mystack; 16 p = root; 17 if(p) mystack.push(p); 18 while(!mystack.empty()) 19 { 20 p = mystack.top(); 21 mystack.pop(); 22 result.push_back(p->val); 23 if(p->right) mystack.push(p->right); 24 if(p->left) mystack.push(p->left); 25 } 26 return result; 27 } 28 };
递归的方法很统一。先序，中序，后续，只要修改遍历的顺序即可。
1 /** 2 * Definition for a binary tree node. 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 public: 12 vector<int> preorderTraversal(TreeNode* root) { 13 vector<int> result; 14 traversal(root,result); 15 return result; 16 } 17 void traversal(TreeNode* root,vector<int>& ret) 18 { 19 if(root) 20 { 21 ret.push_back(root->val); 22 traversal(root->left,ret); 23 traversal(root->right,ret); 24 } 25 } 26 };
[LeetCode]Binary Tree Preorder Traversal
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原文地址：http://www.cnblogs.com/Sean-le/p/4782525.html
