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algorithm/Week_02/id_131/LeetCode_783_131.cpp at master · feixiangcode/algorithm · GitHub
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LeetCode_783_131.cpp
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/*
题目描述 :
* 给定一个二叉搜索树的根结点 root, 返回树中任意两节点的差的最小值。
*/
/*
*
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/*
思路 :
* 既然是二叉搜索树，根据树的性质，知道root的右子树都是大于它的，左子树都是小于它的
* 那么如果做中序遍历，标准的做法是得到一个递增的序列
* 我们就先遍历右根，节点，左根，这样就会得到一个递减的序列。
* 然后对这个序列相邻相减，取最小值即可。 实现时，可以优化掉这个序列。
* 在遍历时记录上一个访问的节点值，和当前节点相减，记录下最小值即可。
* 这样就可以做到时间复杂度O(n)和空间复杂度O(1)
*/
class
Solution
{
int
pre
=
INT_MIN
;
int
minVal =
INT_MAX
;
public:
void
recycle
(TreeNode* root)
{
if
( root ==
NULL
)
return
;
recycle
(root->
left
);
if
(
INT_MIN
!=
pre
)
{
minVal = minVal < (root->
val
-
pre
) ? minVal : (root->
val
-
pre
);
}
pre
= root->
val
;
recycle
(root->
right
);
}
int
minDiffInBST
(TreeNode* root)
{
if
(
NULL
== root)
return
0
;
recycle
(root);
return
minVal;
}
};
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