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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?
题意：先序遍历
思路：采用递归实现。因为函数声明是返回一个vector<int>,所以每个子树返回的是该子树的先序遍历的结果
按照 根、左、右的次序把根和左右子树的vector合并起来就可以了
复杂度：时间O(n)，空间O(n)
相关题目：
Binary Tree Postorder Traversal
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
vector<int> pre;
if(root == NULL) return pre;
TreeNode *left = root->left;
TreeNode *right = root->right;
pre.push_back(root->val);
if(left) {
vector<int> left_vector = preorderTraversal(left);
pre.insert(pre.end(), left_vector.begin(), left_vector.end());
}
if(right){
vector<int> right_vector = preorderTraversal(right);
pre.insert(pre.end(), right_vector.begin(), right_vector.end());
}
return pre;
}
};Leetcode 树 Binary Tree Preorder Traversal,布布扣,bubuko.com
Leetcode 树 Binary Tree Preorder Traversal
原文：http://blog.csdn.net/zhengsenlie/article/details/25722921