Flatten Binary Tree to Linked List

LeetCode #114

Description:

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

Example:

Given

         1
        / \
       2   5
      / \   \
     3   4   6
The flattened tree should look like:
   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6

Idea:

recursion思想很复杂，如果用stack来iteration做。

push的时候先push右边，再push左边。然后立马把node->left=nullptr，right subtree就应该是刚push进stack的元素（如果stack not empty).

Code:

class Solution {
public:
    void flatten(TreeNode* root) {
        stack<TreeNode *> stack_node;
        TreeNode *node=root;

        if(root!=NULL) stack_node.push(root);

        while(!stack_node.empty()){
            node=stack_node.top();
            stack_node.pop(); // after pop, stack could be empty.

            // Push right subtree first, then left subtree!
            if(node->right!=NULL) stack_node.push(node->right);
            if(node->left!=NULL) stack_node.push(node->left);

            node->left=NULL;
            if(!stack_node.empty()){
                node->right=stack_node.top();
            }
        }

    }
};