Leetcode 94 Binary Tree inorder Traversal

Given a binary tree, return the inorder traversal of its nodes ' values.

For example:
Given binary Tree {1,#,2,3} ,

   1         2    /   3

Return [1,3,2] .

Note:recursive solution is trivial, could do it iteratively?

Note that In-order is the root traversal

1. If you use recursion on the left root right

1 classSolution {2  Public:3vector<int> Inordertraversal (treenode*root) {4vector<int>v;5         if(Root = =NULL)6             returnv;7         if(Root->left = = NULL && root->right = =NULL)8         {9V.push_back (root->val);Ten             returnv; One         } Avector<int> left = inordertraversal (root->Left ); -vector<int> right = Inordertraversal (root->Right ); -  the V.insert (V.end (), Left.begin (), Left.end ()); -V.push_back (root->val); - V.insert (V.end (), Right.begin (), Right.end ()); -  +         returnv; -     } +};

2. Non-recursive algorithm:

A first put the left subtree all in, and then start to play, access to it, and then put its right sub-tree pressure in, actually started to bounce out of time, that point can be regarded as a child, can also be seen as the root. So it's left-"root" right

1 classSolution {2  Public:3vector<int> Inordertraversal (treenode*root) {4vector<int>v;5         if(Root = =NULL)6             returnv;7TreeNode * p =Root;8Stack<treenode*>s;9          while(P)Ten         { One              while(p)//keep the tree on the left in the press. A             { - S.push (p); -p = p->Left ; the             } -              while(!s.empty ())//Pop a dot to see if there is no right child, there are words to take out as root to traverse -             { -TreeNode * cur =s.top (); + S.pop (); -V.push_back (cur->val); +                 if(cur->Right ) A                 { atp = cur->Right ; -                      Break; -                 } -             } -         } -         returnv; in     } -};

B with a bit to mark, the root has not been traversed. If it was 0, it was not traversed, and his right child and his left child were put in. If 1 is traversed, then the output is processed directly.

This method is relatively simple, but also faster, recommended memory.

1 classSolution {2  Public:3vector<int> Inordertraversal (treenode*root) {4std::vector<int>v;5         if(Root = =NULL)6             returnv;7         if(Root->left = = NULL && root-right = =NULL) {8V.push_back (root->val);9             returnv;Ten         }  OneStack <pair<treenode*,int> >s; AS.push (Make_pair (Root,0)); -          while(!s.empty ()) -         { theTreeNode * p =S.top (). First; -             intindex =S.top (). Second; - S.pop (); -             if(!index) +             { -                 if(p->Right ) +S.push (Make_pair (P->right,0)); AS.push (Make_pair (P,1)); at                 if(p->Left ) -S.push (Make_pair (P->left,0)); -}Else { -V.push_back (p->val); -             } -         }  in         returnv; -     } to};

Leetcode 94 Binary Tree inorder Traversal