给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
/**
* 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> postorderTraversal(TreeNode* root) {
vector<int> res;
if (!root){
return res;
}
stack<TreeNode*> stack;
TreeNode* tmp = root;
stack.push(tmp);
stack.push(tmp);
while(!stack.empty())
{
tmp = stack.top();
stack.pop();
if(!stack.empty() && tmp == stack.top())
{
if(tmp->right)
{
stack.push(tmp->right);
stack.push(tmp->right);
}
if(tmp->left)
{
stack.push(tmp->left);
stack.push(tmp->left);
}
}
else{
res.push_back(tmp->val);
}
}
return res;
}
};