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?
思路：用栈。之间复杂度O(n),空间复杂度O(n)
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 public: 12 vector<int> preorderTraversal(TreeNode *root) { 13 vector<int> result; 14 stack<TreeNode*> s; 15 TreeNode *p = root; 16 17 while (p != NULL || !s.empty()) { 18 if (p != NULL) { 19 result.push_back(p->val); 20 s.push(p); 21 p = p->left; 22 } else { 23 p = s.top(); 24 s.pop(); 25 p = p->right; 26 } 27 } 28 return result; 29 } 30 };
[LeetCode] Binary Tree Preorder Traversal
原文：http://www.cnblogs.com/vincently/p/4229654.html