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?
OJ's Binary Tree Serialization: The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
递归方法
void recursive(vector<int> &result, TreeNode *root) {
if (root == nullptr)
return;
recursive(result, root->left);
result.push_back(root->val);
recursive(result, root->right);
}迭代方法
void loop(vector<int> &result, TreeNode *root) {
if (root == nullptr)
return;
stack<TreeNode *> s;
TreeNode *cur = root;
while (cur || !s.empty()) {
while (cur) {
s.push(cur);
cur = cur->left;
}
cur = s.top();
s.pop();
result.push_back(cur->val);
cur = cur->right;
}
}