【数据结构_二叉树+dfs】

Flatten Binary Tree to Linked List

 Total Accepted: 17814 Total Submissions: 64054My Submissions

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

/** * 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) {        dfs(root);    }    TreeNode *dfs(TreeNode *root){        if(root==NULL) return NULL;        TreeNode *left=root->left,*leftTail=dfs(root->left),                 *right=root->right,*rightTail=dfs(root->right);        if(left) {            root->right=left,root->left=NULL;            if(right){                leftTail->right=right;                return rightTail;            }else{                return leftTail;            }         }else{//left child is null            if(right) return rightTail;            else return root;//left and right are null        }    }};

Validate Binary Search Tree

 Total Accepted: 16968 Total Submissions: 66123My Submissions

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

class Solution {public:    bool isValidBST(TreeNode *root) {        if(root==NULL) return true;        Pr tmp;return dfs(root,tmp);    }    typedef pair<int,int> Pr;    bool dfs(TreeNode *root,Pr &pr){      assert(root!=NULL);      int _min=root->val,_max=root->val;      if(root->left) {          Pr p1;          if(!dfs(root->left,p1)) return false;          if(p1.second>=root->val) return false; //@@error: pair p1; p1->second should be p1.second.//@error:p1.second>root->val,should be >=          _min=min(_min,p1.first);      }      if(root->right) {          Pr p2;          if(!dfs(root->right,p2)) return false;          if(p2.first<=root->val) return false;          _max=max(_max,p2.second);      }      pr=Pr(_min,_max);      return true;    }};

或者

为dfs(Node * root,int l,int r)//检验root是否在[l,r]范围内，不是则不合格

class Solution {public:#define MIN 0x80000000#define MAX 0x7FFFFFFF    bool isValidBST(TreeNode *root) {        return dfs(root,MIN,MAX);    }   bool dfs(TreeNode *root,int l,int r){       if(root==NULL) return true;       int val=root->val;       if(val<l||val>r) return false;       return dfs(root->left,l,val-1)&&dfs(root->right,val+1,r);   }};