114. Flatten Binary Tree to Linked List (Stack, Tree; DFS)

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

For example,
Given

         1
        /        2   5
      / \        3   4   6

The flattened tree should look like:

   1
         2
             3
                 4
                     5
                         6
法I：堆栈

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) return;
        TreeNode* flatRoot = new TreeNode(root->val);
        TreeNode* flatNode = flatRoot;
        TreeNode* current;
        stack<TreeNode*> tree_stack;

        //栈是先入后出，所以反序放入栈，即先右儿子，再左儿子
        if(root->right)
        {
            tree_stack.push(root->right);
        }

        if(root->left)
        {
            tree_stack.push(root->left);
        }
        while(!tree_stack.empty())
        {
            current = tree_stack.top();
            tree_stack.pop();

            flatNode->right = new TreeNode(current->val);
            flatNode = flatNode->right;

            //将出栈元素视为根节点，再次将它的右儿子、左儿子放入栈
            if(current->right)
            {
                tree_stack.push(current->right);
            }

            if(current->left)
            {
                tree_stack.push(current->left);
            }
        }

        *root = *flatRoot;
    }
};

法II：递归，前序遍历

class Solution {
public:
    void flatten(TreeNode *root) {
        if(!root) return;
        TreeNode* newRoot = new TreeNode(root->val);
        preOrderTraverse(root, newRoot);
        *root = *newRoot;
    }
    TreeNode* preOrderTraverse(TreeNode *root, TreeNode * newRoot)
    {
        if(root->left)
        {
            newRoot->right = new TreeNode(root->left->val);
            newRoot = preOrderTraverse(root->left, newRoot->right);
        }
        if(root->right)
        {
            newRoot->right = new TreeNode(root->right->val);
            newRoot = preOrderTraverse(root->right, newRoot->right);
        }
        return newRoot;
    }
};