LeetCode 题解(266) : Inorder Successor in BST

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题目：

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

Note: If the given node has no in-order successor in the tree, return null.

题解：

用了递归。循环应该略有难度。

C++版

/** 
* Definition for a binary tree node. 
* struct TreeNode { 
*     int val; 
*     TreeNode *left; 
*     TreeNode *right; 
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 
* }; 
*/ 
class Solution { 
public: 
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) { 
inorder(root, p); 
return successor; 
} 
bool inorder(TreeNode* root, TreeNode* p) { 
if(root->left) { 
if(inorder(root->left, p)) 
return true; 
} 
if(found) { 
successor = root; 
return true; 
} 
if(root == p) 
found = true; 
if(root->right) { 
if(inorder(root->right, p)) 
return true; 
} 
return false; 
} 
TreeNode* successor = NULL; 
bool found = false; 
};

Python版：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def __init__(self):self.find = Falseself.successor = Nonedef inorderSuccessor(self, root, p):""":type root: TreeNode:type p: TreeNode:rtype: TreeNode"""self.inorder(root, p)return self.successordef inorder(self, root, p):if root.left != None:if self.inorder(root.left, p):return Trueif self.find:self.successor = rootreturn Trueif root == p:self.find = Trueif root.right != None:if self.inorder(root.right, p):return Truereturn False

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