Leetcode -- flatten binary tree to linked list

Problem description:

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

1/2 5/\ 3 4 6
 

The flattened tree shoshould look like:

1 2 3 4 5 6
 

Click to show hints.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

Analysis: According to the meaning of the question, the binary tree is changed to an exclusive right subtree arranged in ascending order. Therefore, the solution is to traverse the binary tree in ascending order and save the node first, then re-build the binary tree as required. In practice, the non-recursive first-order traversal is used to save the node to the vector, and finally the re-build the binary tree. The Code is as follows:

/*** Definition for binary tree * struct treenode {* int val; * treenode * left; * treenode * right; * treenode (int x): Val (x ), left (null), right (null) {}*}; */class solution {public: void flatten (treenode * root) {If (root = NULL) return; vector <treenode *> VEC; stack <treenode *> STK; treenode * P = root; while (p |! STK. Empty () {While (p) {Vec. push_back (p); STK. Push (p); P = p-> left;} If (! STK. empty () {P = STK. top (); STK. pop (); P = p-> right ;}} root = VEC [0]; for (INT I = 0; I <Vec. size ()-1; I ++) {VEC [I]-> left = NULL; VEC [I]-> right = VEC [I + 1];} VEC [Vec. size ()-1]-> left = NULL; VEC [Vec. size ()-1]-> right = NULL ;}};