LeetCode 114. Flatten Binary Tree to Linked List 动态演示
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把二叉树先序遍历,变成一个链表,链表的next指针用right代替
用递归的办法先序遍历,递归函数要返回子树变成链表之后的最后一个元素
class Solution public: void helper(TreeNode* cur, TreeNode*& tail) //a(tail) //lk("root",tail) //a(cur) //lk("root",cur) //dsp tail=cur; TreeNode* right=cur->right; //a(right) //lk("root",right) if(cur->left) TreeNode *leftTail=NULL; helper(cur->left, leftTail); cur->right=cur->left; cur->left=NULL; tail=leftTail; //dsp if(right) TreeNode *rightTail=NULL; helper(right, rightTail); tail->right=right; tail=rightTail; //dsp void flatten(TreeNode* root) if(!root) return; //ahd(root) TreeNode *tail=NULL; helper(root, tail); ;
程序运行动态演示:http://simpledsp.com/FS/Html/lc114.html
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