145. Binary Tree Postorder Traversal
Total Accepted: 96378 Total Submissions: 271797 Difficulty: Hard
提交网址： https://leetcode.com/problems/binary-tree-postorder-traversal/
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?
分析：
AC代码：
#include<iostream>
#include<vector>
#include<stack>
#include<algorithm> // 引入 reverse函数，对vector进行反转
using namespace std;
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(0);
if(root==NULL) return res;
stack<TreeNode *> st;
TreeNode *p=root;
while(!st.empty()|| p!=NULL)
{
if(p!=NULL)
{
st.push(p);
res.push_back(p->val);
p=p->right;
}
if(p==NULL)
{
p=st.top();
p=p->left;
st.pop();
}
}
reverse(res.begin(),res.end()); // 对向量及其内部结构进行反转
return res;
}
};
// 以下为测试
int main()
{
Solution sol;
vector<int> res;
TreeNode *root = new TreeNode(1);
root->right = new TreeNode(2);
root->right->left = new TreeNode(3);
res=sol.postorderTraversal(root);
for(int i:res)
cout<<i<<" "; // 此处为vector遍历的方法，C++11标准支持
return 0;
}另一解法(非递归)：
后序遍历的麻烦在于访问过右子树之后，第二次访问的时候就必须访问根节点，而不是继续访问右子树，所以使用pre来记录上次访问的节点...
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> ans;
if(root == NULL) return ans;
stack<TreeNode* > s;
TreeNode * p = root;
TreeNode * pre = NULL; // 记录上次访问的节点
while(p != NULL || !s.empty()) {
while(p != NULL) {
s.push(p);
p = p->left;
}
if(!s.empty()) {
p = s.top();
s.pop();
if(p->right == NULL || p->right == pre) { // 右子树为空，或者是访问过
ans.push_back(p->val);
pre = p; // 记录刚刚访问的右节点
p = NULL;
}
else {
s.push(p);
p = p->right;
}
}
}
return ans;
}
};