dfs O(n)
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class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
int res = 0;
dfs(root, k, res);
return res;
}
void dfs(TreeNode *root, int &k, int &res) {
if (!root) return;
dfs(root->left, k, res);
if (--k == 0) {
res = root->val;
return;
}
dfs(root->right, k, res);
}
};
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follow up 给每个节点加一个count来记录当前节点这棵树的所有节点的个数，初始化为1，即只有根节点
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class Solution {
public:
int kthSmallest(TreeNode* root, int k) {
TreeNodeWithCount *node = buildTree(root);
return kthSmallest(node, k);
}
private:
struct TreeNodeWithCount {
int val;
int count;
TreeNodeWithCount *left;
TreeNodeWithCount *right;
TreeNodeWithCount(int x) : val(x), count(1), left(nullptr), right(nullptr) {}
};
TreeNodeWithCount * buildTree(TreeNode *root) {
if (!root)
return nullptr;
TreeNodeWithCount *node = new TreeNodeWithCount(root->val);
node->left = buildTree(root->left);
node->right = buildTree(root->right);
if (node->left)
node->count += node->left->count;
if (node->right)
node->count += node->right->count;
return node;
}
int kthSmallest(TreeNodeWithCount *root, int k) {
if (root->left) {
if (root->left->count >= k)
return kthSmallest(root->left, k);
if (root->left->count == k - 1)
return root->val;
return kthSmallest(root->right, k - 1 - root->left->count);
}
else {
if (k == 1)
return root->val;
return kthSmallest(root->right, k - 1);
}
}
};
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