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?
用递归时, 原先给的函数不便处理, 需要转化一下
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void preorder(TreeNode *root, vector<int> &temp) { if(!root) return ; temp.push_back(root->val); preorder(root->left,temp); preorder(root->right,temp); } vector<int> preorderTraversal(TreeNode *root) { vector<int> result; preorder(root, result); return result; } };