Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively
一次通过~~ 耶
/**
* 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 inorderTraversal(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(!root) return vector();
vector result;
stack> s;
s.push(pair(root, false));
TreeNode *curNode;
bool curFlag;
while(!s.empty()){
curNode = s.top().first;
curFlag = s.top().second;
s.pop();
if(curFlag){
//second access
result.push_back(curNode->val);
}else{
//first access, push right, self, left
if(curNode -> right) s.push(pair(curNode->right, false));
s.push(pair(curNode, true));
if(curNode -> left) s.push(pair(curNode->left, false));
}
}
return result;
}
};