[Locked] Inorder Successor in BST

Inorder Successor in BST

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

Example

Given tree =

[2,1]

and node =

1

:

2
/
1

return node

2

.

Given tree =

[2,1,3]

and node =

2

:

2
/ \
1   3

return node

3

.

Note

If the given node has no in-order successor in the tree, return

null

.

Challenge

O(h), where h is the height of the BST.

分析：

　　一般情况下，目标节点的右子节点即为中序下一个节点；如果该目标节点无右子节点，则中序下一个节点为从根节点到搜索目标节点过程中最后一次做出左节点遍历的节点。

代码：

TreeNode *successor(TreeNode *root, TreeNode *p) {
TreeNode *cur = root, *record = NULL;
while(cur != p) {
if(p->val < cur->val) {
record = cur;
cur = cur->left;
}
else
cur = cur->right;
}
return cur->right ? cur->right : record;
}