剑指offer:二叉搜索树的第k个结点

递归

/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
*/
class Solution {
public:
TreeNode* KthNode(TreeNode* pRoot, unsigned int k)
{
if (pRoot == NULL || k == 0){
return NULL;
}

return helper(pRoot, k);
}

private:
TreeNode* helper(TreeNode* pRoot, unsigned int& k){
TreeNode* res = NULL;

if (pRoot->left){
//
res=helper(pRoot->left, k);
}

if (pRoot){
if (res == NULL){
if (k == 1){
res = pRoot;
}
k--;
}
}

//res==NULL就不用再递归
if (res==NULL && pRoot->right!=NULL){
//
res=helper(pRoot->right, k);
}

return res;
}
};

非递归

/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
*/
class Solution {
public:
TreeNode* KthNode(TreeNode* pRoot, unsigned int k)
{
if (pRoot == NULL || k == 0){
return NULL;
}

stack<TreeNode*> stk;
TreeNode* p = pRoot;

int index = 0;
TreeNode* res=NULL;

while (p != NULL || !stk.empty()){
while (p){
stk.push(p);
p = p->left;
}

//这里是if，不是while
if (!stk.empty()){
p = stk.top();
stk.pop();

index++;

if (index == k){
res = p;
break;
}

//不用判断p->right是否为空
p = p->right;
}

}

return res;
}
};