LeetCode -- 二叉树的后序遍历（最详细的解法！！！）
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？
解题思路：总体思路分为递归和迭代两种。两者都是依赖于后序遍历的顺序：左、右、根。具体细节看下面代码。
代码一：
/*递归1、最简单的算法*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
if(root){
postorderTraversal(root->left);
postorderTraversal(root->right);
res.push_back(root->val);
}
return res;
}
private:
vector res;
};*/
代码二：
/*递归2*/
/*class Solution {
public:
vector postorderTraversal(TreeNode *root) {
vector res;
stack res_;
postorder(root, res_);
while(!res_.empty()){
res.push_back(res_.top());
res_.pop();
}
return res;
}
void postorder(TreeNode *root, stack &res) {
if (!root)
return;
res.push(root->val);
if (root->right)
postorder(root->right, res);
if (root->left)
postorder(root->left, res);
}
};*/
代码三：
/*递归3*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
vector res;
postorder(root, res);
return res;
}
void postorder(TreeNode *root, vector &res) {
if (!root)
return;
if (root->left)
postorder(root->left, res);
if (root->right)
postorder(root->right, res);
res.push_back(root->val);
}
};*/
代码四：
/*迭代1*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
vector res;
vector res_;
stack s;
TreeNode *p = root;
if(p == NULL)
return res;
while(p || !s.empty()){
if(p != NULL){
res.push_back(p->val);
s.push(p->left);
p = p->right;
}
else{
p = s.top();
s.pop();
}
}
for(int i = res.size()-1;i >= 0;--i)
res_.push_back(res[i]);
return res_;
}
};*/
代码五：
/*迭代2*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
stack s;
vector res;
if(root==NULL)
return res;
TreeNode* p = root;
TreeNode* temp = NULL;
while(p||!s.empty()){
if(!p){
p=s.top();//弹出栈内结点
s.pop();
if(p->right){//此节点有右子树，P转到该右子树，并把该结点的右子树置为NULL，押回
temp=p->right;
p->right=NULL;
s.push(p);
p=temp;
}
else{
res.push_back(p->val);
p=NULL;
}
}
else{
s.push(p);
p=p->left;
}
}
return res;
}
};*/
代码六：
/*迭代3*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
vector res;
stack s;
TreeNode *p = root;
while (!s.empty() || p) {
if (p) {
s.push(p);
res.insert(res.begin(), p->val); //反向添加，而前序是正向添加
p = p->right;
}
else {
TreeNode *t = s.top();
s.pop();
p = t->left;
}
}
return res;
}
};*/
代码七：
/*迭代4*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
vector res;
stack s;
TreeNode *p = root;
while(!s.empty() || p){
if(p){
s.push(p);
p = p->left;
}
else{
if(p->right== NULL ){
res.push_back(p->val);
s.pop();
}
else
p = p->right;
}
}
return res;
}
};*/
代码八：
/*迭代5*/
//常数时间空间复杂度
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
vector res;
TreeNode *tmp = NULL;
while(root){
tmp=root->right;
while(tmp && tmp->left!=NULL && tmp->left!=root){
tmp = tmp->left;
}
if(tmp == NULL){
res.push_back(root->val);
root=root->left;
}
else if(tmp->left == NULL){
res.push_back(root->val);
tmp->left=root;
root=root->right;
}
else{
tmp->left=NULL;
root=root->left;
}
}
reverse(res.begin(),res.end());
return res;
}
};*/
代码九：
/*迭代6*/
/*class Solution {
public:
vector postorderTraversal(TreeNode* root) {
vector ans;
if(!root)
return ans;
stack st;
TreeNode *p = root;
st.push(p);
while(!st.empty()){
p = st.top();
if(!p->left && !p->right){
st.pop();
ans.push_back(p->val);
}
if(p->right){
st.push(p->right);
p->right = NULL;
}
if(p->left){
st.push(p->left);
p->left = NULL;
}
}
return ans;
}
};*/