LeetCode(114)Flatten Binary Tree to Linked List

题目分析

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.

分析如下:

因为最后flatten的结果是树的前序遍历的结果，所以考虑一边进行前序遍历，一边进行flatten转化.

代码如下:

//48ms过大集合/** * Definition for binary tree * 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==NULL)            return;        stack<TreeNode*> node_stack;        node_stack.push(root);        TreeNode* new_root=NULL;        TreeNode* cur=NULL;        TreeNode* next=NULL;        while(!node_stack.empty()){            cur=node_stack.top();            node_stack.pop();            if(cur->right!=NULL)                node_stack.push(cur->right);            if(cur->left!=NULL)                node_stack.push(cur->left);            if(new_root==NULL){                new_root=cur;                next=cur;                cur->left=NULL;            } else {                next->right=cur;                next->left=NULL;                next=cur;            }        }        root=new_root;    }};

小结:

(1) 逻辑很重要，在while循环体中，应该先把cur->right, cur->left压栈，再去进行flatten。如果颠倒了顺序，就会在flatten时破坏一些还没有被处理的节点，这些节点被压栈，随后就会发生错误。