5.1.9 Flatten Binary Tree to Linked List

Notes:
	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
	Hints:
	If you notice carefully in the flattened tree, each node's right child points to the next node
	of a pre-order traversal.
	Solution: Recursion. Return the last element of the flattened sub-tree.
	*/

/**
	* Definition for binary tree
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/

思路：
利用栈：现将右接点压入栈，再将左节点压入栈，在弹出来的时候，左节点赋NULL，有节点指向栈顶元素

class Solution {
public:
    void flatten(TreeNode *root) {
        flatten_3(root);
    }

    void flatten_1(TreeNode *root) {
        if (root == NULL) return;
        flatten_1(root->left);
        flatten_2(root->right);
        if (root->left == NULL) return;
        TreeNode *p = root->left;
        while (p->right) p = p->right;
        p->right = root->right;
        root->right = root->left;
        root->left = NULL;
    }

    TreeNode * dfs (TreeNode * root, TreeNode * tail){
        if(root == NULL) return tail;
        root->right = dfs(root->left,dfs(root->right,tail));
        root->left = NULL;
        return root;
    }
    void flatten_2(TreeNode *root) {
        if(root == NULL) return;
        dfs(root, NULL);
    }
    void flatten_3(TreeNode *root) {
        if(root==nullptr) return;
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty()){
            auto p=s.top();
            s.pop();
            if(p->right) s.push(p->right);
            if(p->left) s.push(p->left);
            p->left = nullptr;
            if(!s.empty()){
                p->right=s.top();
            }else p->right = nullptr;
        }
    }

    void flatten_4(TreeNode *root) {
        TreeNode *end = NULL;
        flattenRe(root, end);
    }

    void flattenRe(TreeNode *node, TreeNode *&end) {
        if (!node) return;
        TreeNode *lend = NULL, *rend = NULL;
        flattenRe(node->left, lend);
        flattenRe(node->right, rend);
        if (node->left) {
            lend->right = node->right;
            node->right = node->left;
            node->left = NULL;
        }
        end = rend ? rend : (lend ? lend : node);
    }
};