原文地址：https://www.jianshu.com/p/68e6562c666d
时间限制：1秒 空间限制：32768K
题目描述
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1
2
/
3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
我的代码
/**
* Definition for binary tree
* 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*> st;
st.push(root);
TreeNode* pre=nullptr;
while(!st.empty()){
TreeNode* cur=st.top();
if(((cur->left==nullptr)&&(cur->right==nullptr))
||(pre!=nullptr&&(pre==cur->left||pre==cur->right))){
res.push_back(cur->val);
st.pop();
pre=cur;
}
else{
if(cur->right){
st.push(cur->right);
}
if(cur->left){
st.push(cur->left);
}
}
}
return res;
}
};
运行时间：3ms
占用内存：476k
/**
* Definition for binary tree
* 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*> st;
st.push(root);
while(!st.empty()){
TreeNode* cur=st.top();st.pop();
res.push_back(cur->val);
//根左右进，根右左出
if(cur->left){
st.push(cur->left);
}
if(cur->right){
st.push(cur->right);
}
}
//倒序
reverse(res.begin(),res.end());
return res;
}
};
运行时间：6ms
占用内存：620k