标签：leetcode
问题描述
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?
解决方案
/**
* 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) {
if( root == nullptr ) {
return { };
}
stack<TreeNode*> nodeStack;
nodeStack.push( root->right );
nodeStack.push( root->left );
vector<int> result{ root->val };
while( !nodeStack.empty() ) {
auto t = nodeStack.top();
nodeStack.pop();
if( t != nullptr ) {
result.push_back( t->val );
nodeStack.push( t->right );
nodeStack.push( t->left );
}
}
return result;
}
};Binary Tree Preorder Traversal
标签：leetcode
原文地址：http://blog.csdn.net/senlinzm/article/details/39008353
