"Binary tree Traversal" in sequence

Record the available middle-order traversal methods

Method 1:

vector<int> Inordertraverse (TreeNode * root)//Middle Sequence Traversal{vector<int>ans; TreeNode*p; Vector<treenode *>Stack;        Stack.push_back (root);  while(!Stack.empty ()) {             while(P = stack.back ())! =NULL) Stack.push_back (P-Left );            Stack.pop_back (); if(!Stack.empty ()) {P=Stack.back ();                Stack.pop_back (); Ans.push_back (P-val); Stack.push_back (P-Right ); }        }        returnans; }

Method 2:

vector<int> Inordertraversal (TreeNode *root) {Vector<int>Vector; Stack<treenode *>Stack; TreeNode*pcurrent =Root;  while(!stack.empty () | |pcurrent) {            if(pcurrent) {Stack.push (pcurrent); Pcurrent= pcurrent->Left ; }            Else{TreeNode*pnode =Stack.top (); Vector.push_back (Pnode-val);                Stack.pop (); Pcurrent= pnode->Right ; }        }        returnVector; }

I wrote it in my own way, and tuned it many times. The key is the logical confusion when the right subtree pops up.

vector<int> Inordertraversal (TreeNode *root) {Vector<int>ans; Vector<treenode *>v; TreeNode*p; if(Root = NULL)returnans;        V.push_back (root);  while(!V.empty ()) {P=V.back ();  while(P->left! = NULL)//find the left-most{v.push_back (P-Left ); P= p->Left ; } ans.push_back (P->val);//Press-in dataV.pop_back ();  while(P->right = = NULL)//right subtree processing if the current right subtree is empty then you need to eject the value of the tail in V            {                if(!V.empty ()) {P=V.back (); Ans.push_back (P-val);                V.pop_back (); }                Else                {                     Break; }            }            if(P->right! =NULL) {V.push_back (P-Right ); }        }        returnans; }

There is no need to stack the method, using the Clue two fork tree concept, Space complexity O (1)

Http://www.cnblogs.com/AnnieKim/archive/2013/06/15/morristraversal.html

Binary tree traversal in sequence