给定一个二叉树的根节点root,返回它的中序遍历 。
示例 1:

输入:root = [1,null,2,3]
输出:[1,3,2]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
提示:
- 树中节点数目在范围
[0, 100]内 -100 <= Node.val <= 100
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
思路:
输出顺序为:左节点->根节点->右节点,使用栈暂存即可
AC代码:
/**
* 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*> nodeStack;
if(root == NULL)
{
return result;
}
while(root || !nodeStack.empty())
{
while(root)
{
nodeStack.push(root);
root = root->left;
}
root = nodeStack.top();
nodeStack.pop();
result.push_back(root->val);
root = root->right;
}
return result;
}
};Code language: C++ (cpp)