题目描述:(链接）
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?
解题思路：
迭代版：
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 stack<TreeNode *> cache; 15 TreeNode *p = root; 16 if (p != nullptr) { 17 cache.push(p); 18 } 19 20 while (!cache.empty()) { 21 p = cache.top(); 22 cache.pop(); 23 result.push_back(p->val); 24 25 if (p->right != nullptr) { 26 cache.push(p->right); 27 } 28 29 if (p->left != nullptr) { 30 cache.push(p->left); 31 } 32 } 33 34 return result; 35 } 36 };
递归版：
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 if (root != nullptr) { 14 result.push_back(root->val); 15 preorderTraversal(root->left); 16 preorderTraversal(root->right); 17 } 18 19 return result; 20 } 21 private: 22 vector<int> result; 23 };
[LeetCode]Binary Tree Preorder Traversal
原文：http://www.cnblogs.com/skycore/p/4986124.html