lintcode-453-将二叉树拆成链表

453-将二叉树拆成链表

将一棵二叉树按照前序遍历拆解成为一个假链表。所谓的假链表是说，用二叉树的 right 指针，来表示链表中的 next 指针。

注意事项

不要忘记将左儿子标记为 null，否则你可能会得到空间溢出或是时间溢出。

样例

挑战

不使用额外的空间耗费。

标签

二叉树 深度优先搜索

方法一

使用栈保存暂时无法插入到链表的节点

code

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /*
     * @param root: a TreeNode, the root of the binary tree
     * @return: 
     */
    void flatten(TreeNode * root) {
        // write your code here
        if (root == NULL) {
            return ;
        }
        stack<TreeNode *>stack;
        TreeNode * node = root;
        bool isContinue = true;
        while (isContinue) {
            if (node->left != NULL && node->right != NULL) {
                stack.push(node->right);
                node->right = node->left;
                node->left = NULL;
            }
            else if (node->left != NULL && node->right == NULL) {
                node->right = node->left;
                node->left = NULL;
            }
            else if (node->left == NULL && node->right == NULL) {
                if (!stack.empty()) {
                    node->right = stack.top();
                    stack.pop();
                }
                else {
                    isContinue = false;
                }
            }
            node = node->right;
        }
    }
};

方法二

使用递归

code

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /*
     * @param root: a TreeNode, the root of the binary tree
     * @return: 
     */
    void flatten(TreeNode * root) {
        // write your code here
        if (root == NULL) {
            return;
        }
        switchToLink(root);
    }

    TreeNode* switchToLink(TreeNode* root) {
        if (root == NULL) {
            return NULL;
        }

        TreeNode* left = switchToLink(root->left);
        TreeNode* right = switchToLink(root->right);

        if (left != NULL) {
            left->right = root->right;
            root->right = root->left;
        }

        root->left = NULL;

        if (right != NULL) {
            return right;
        }
        else if (left != NULL) {
            return left;
        }
        else {
            return root;
        }
    }
};