[LeetCode] Binary Tree Inorder Traversal

Binary Tree Inorder Traversal

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?

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

解题思路：

中序遍历，递归算法没有啥问题，先左子树，然后根节点，然后右子树。

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
helper(root, result);
return result;
}

void helper(TreeNode* root, vector<int>& result){
if(root == NULL) {
return;
}
helper(root->left, result);
result.push_back(root->val);
helper(root->right, result);
}
};

迭代版本：

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> s;
TreeNode* node = root;
while(!s.empty() || node!=NULL){
while(node!=NULL){
s.push(node);
node = node->left;
}
node = s.top();
s.pop();
result.push_back(node->val);
node = node->right;
}

return result;
}
};