标签：des blog http os io strong for ar
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?
1st （ 2 tries）
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//help from http://www.cnblogs.com/dolphin0520/archive/2011/08/25/2153720.html
//there is all kind all traversal methods with binary tree in this
class Solution
{
public:
vector<int> postorderTraversal(TreeNode *root)
{
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
TreeNode *iter = NULL;
TreeNode *preiter = NULL;
vector<int> re;
stack<TreeNode*> st;
if(root == NULL)
return re;
st.push(root);
while(!st.empty())
{
iter = st.top();
if(iter->left == NULL&&iter->right == NULL)
{
re.push_back(iter->val);
preiter = iter;
st.pop();
}
else if(preiter != NULL && (preiter == iter->left || preiter == iter->right))
{
re.push_back(iter->val);
preiter = iter;
st.pop();
}
else
{
if(iter->right != NULL)
st.push(iter->right);
if(iter->left != NULL)
st.push(iter->left);
}
}
return re;
}
};
2nd （15 tries)
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
//learn morris post-order traversal!!! this is most difficult!!!
vector<int> ans;
vector<int> postorderTraversal(TreeNode *root) {
//now I can‘t get the iterative postorder traversal
//rethink!!!
//now I just remember preorder traversal
//preorder and inorder are not so differcut,just change ans.push_back() position!!!
//postorder???
/*
stack<TreeNode*> s;
vector<TreeNode*> roots;
TreeNode *cur = root;
while(!s.empty() || cur != NULL) {
while(cur != NULL) {
s.push(cur);
cur = cur->left;
}
cur = s.top();
//now the root have out of stack,how to save it???
//This method change Tree structure???how to don‘t change tree structure!!!
//using flag is one method!!!
if(cur->right == NULL) {
ans.push_back(cur->val);
s.pop();
cur = cur->right;
}
else {
TreeNode *pre = cur;
cur = cur->right;
pre->right = NULL;
}
}
return ans;
*/
//use one stack to obtain postorderTraversal
//root , right, left then reverse the vector!!!
//!!!
/*
stack<TreeNode*> s;
stack<TreeNode*> post;
TreeNode *cur = root;
while( !s.empty() || cur != NULL ) {
while(cur != NULL ) {
ans.push_back(cur->val);
s.push(cur);
cur = cur->right;
}
cur = s.top();
s.pop();
cur = cur->left;
}
vector<int> res;
while( !post.empty() ) {
ans.push_back(post.top()->val);
post.pop();
}
reverse(ans.begin(),ans.end());
return ans;
*/
//another method is using prev!!!
//link:http://leetcode.com/2010/10/binary-tree-post-order-traversal.html
if(root == NULL)
return ans;
stack<TreeNode*> s;
s.push(root);
TreeNode *cur = NULL,*pre = NULL;
while( !s.empty() ) {
cur = s.top();
if( pre == NULL || pre->left == cur || pre->right == cur ) {
if(cur->left != NULL)
s.push(cur->left);
else if(cur->right != NULL)
s.push(cur->right);
}
else if( cur->left == pre ) {
if( cur->right != NULL )
s.push(cur->right);
}
else {
ans.push_back(cur->val);
s.pop();
}
pre = cur;
}
return ans;
}
/*
void dfs(TreeNode *root) {
if(root == NULL)
return;
dfs(root->left);
dfs(root->right);
ans.push_back(root->val);
}
*/
};
/**
* 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> ans;
void reverse(TreeNode *from,TreeNode *to) {
TreeNode *cur = from->left;
TreeNode *post = cur->right;
//reverse covery like reverse link list...
while( cur != to ) {
TreeNode *post_right = post->right;
post->right = cur;
cur = post;
post = post_right;
}
while( cur != from->left ) {
ans.push_back(cur->val);
cur = cur->right;
}
ans.push_back(cur->val);
}
void reversecover(TreeNode *from,TreeNode *to) {
TreeNode *cur = from;
TreeNode *post = cur->right;
TreeNode *end = to->left;
//just reverse like reverse link list
while( cur != end ) {
TreeNode *post_right = post->right;
post->right = cur;
cur = post;
post = post_right;
}
from->right = to;
}
vector<int> postorderTraversal(TreeNode *root) {
//morris
TreeNode *newroot = new TreeNode(0);
newroot->left = root;
newroot->right = NULL;
TreeNode *cur = newroot,*pre = NULL;
while( cur != NULL ) {
if( cur->left == NULL ) {
cur = cur->right;
}
else {
pre = cur->left;
while( pre->right != NULL && pre->right != cur ) {
pre = pre->right;
}
if( pre->right == NULL ) {
pre->right = cur;
cur = cur->left;
}
else {
//post-order here
//reverse
reverse(cur,pre);
//reverse-cover
reversecover(pre,cur);
//over
pre->right = NULL;
cur = cur->right;
}
}
}
return ans;
}
};
【leetcode】Binary Tree Postorder Traversal,布布扣,bubuko.com
【leetcode】Binary Tree Postorder Traversal
标签：des blog http os io strong for ar
原文地址：http://www.cnblogs.com/weixliu/p/3924390.html
