给定一棵二叉树，返回其节点值的后序遍历。
例如：
给定二叉树 [1,null,2,3]，
1
\
2
/
3
返回 [3,2,1]。
注意: 递归方法很简单，你可以使用迭代方法来解决吗？
详见：https://leetcode.com/problems/binary-tree-postorder-traversal/description/
方法一：递归
/**
* 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==nullptr)
{
return res;
}
return postorderTraversalHelper(root,res);
}
vector<int> postorderTraversalHelper(TreeNode* root,vector<int> &res)
{
if(root==nullptr)
{
return res;
}
postorderTraversalHelper(root->left,res);
postorderTraversalHelper(root->right,res);
res.push_back(root->val);
return res;
}
};
方法二：非递归
/**
* 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==nullptr)
{
return res;
}
stack<TreeNode*> stk1;
stack<TreeNode*> stk2;
stk1.push(root);
while(!stk1.empty())
{
root=stk1.top();
stk1.pop();
stk2.push(root);
if(root->left)
{
stk1.push(root->left);
}
if(root->right)
{
stk1.push(root->right);
}
}
while(!stk2.empty())
{
root=stk2.top();
stk2.pop();
res.push_back(root->val);
}
return res;
}
};
