LeetCode: Flatten Binary Tree to Linked List [114]

【题目】

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

click to show hints.

【题意】

    将一颗二叉树转化为链表，right充当next指针，元素顺序为先序遍历的循序。不使用额外的空间

【思路】

    先对左子树链表化，然后将右子树链表化。然后按照:根节点->左子树链表->右子树链表 的顺序组合成完整的链表

【代码】

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* flattenTree(TreeNode *root){        if(root->left==NULL && root->right==NULL)return root;        TreeNode*tail=root;        TreeNode*headLeft=root->left;        TreeNode*tailLeft=NULL;        TreeNode*headRight=root->right;        TreeNode*tailRight=NULL;        //【注意】左孩子清NULL        root->left=NULL;        //将左子树链表化        if(headLeft!=NULL){            tailLeft=flattenTree(headLeft);            tail->right=headLeft;            tail=tailLeft;        }        //将右子树链表化        if(headRight!=NULL){            tailRight=flattenTree(headRight);            tail->right=headRight;            tail=tailRight;        }        return tail;    }        void flatten(TreeNode *root) {        if(root==NULL)return;        flattenTree(root);    }};