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?
/**
* 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> result;
stack<TreeNode*> st;
TreeNode*node=root;
while(node){
result.push_back(node->val);
st.push(node);
node=node->left;
}
while(!st.empty()){
node=st.top()->right; st.pop();
while(node){
result.push_back(node->val);
st.push(node);
node=node->left;
}
}
return result;
}
};LeetCode: Binary Tree Preorder Traversal [144],布布扣,bubuko.com
LeetCode: Binary Tree Preorder Traversal [144]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/35780357
