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

null

.
inorder 遍历 BST 用stack模拟，先把左节点都放进去，在看有没有右节点，如果有再把右节点的左节点都放进去。/**
* 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) {
stack<TreeNode *> s;
if (root == NULL || p == NULL) return NULL;
TreeNode* cur = root;
while (cur != NULL) {
s.push(cur);
cur = cur->left;
}
bool flag = false;

while (!s.empty()) {
cur = s.top();
s.pop();

if (flag == true)
return cur;

if (p->val == cur->val) flag = true;
if (cur->right == NULL) continue; //记得为NULL就continue 之前用不为NULL设置cur=cur->right,出了bag
cur = cur->right;
while(cur != NULL) {
s.push(cur);
cur = cur->left;
}
}

return NULL;
}
};