本來以爲 既然都是找後繼 都是一樣的
結果 都!不!一!樣!
查找的時候怎麼用val比較呢?
考慮如果當前root的值比查找的 小於或等於:淘汰掉左子樹就好
要是需要找到左子樹了,那麼遞歸下去同樣的方法,如果找不到,那麼說明當前這個根節點是要找的
class Solution {
public:
/*
* @param root: The root of the BST.
* @param p: You need find the successor node of p.
* @return: Successor of p.
*/
TreeNode * inorderSuccessor(TreeNode * root, TreeNode * p) {
if (root == NULL) return NULL;
if (root->val <= p->val) {
return inorderSuccessor(root->right, p);
} else {
TreeNode* left = inorderSuccessor(root->left, p);
return left?left:root;
}
}
};
個人喜歡這個非遞歸的 ,先根據p的val找到等於val的節點
然後,根據next的性質,那麼是父節點,那麼是右左左左左
父節點在上次找=val的時候找到的更新
class Solution {
public:
/*
* @param root: The root of the BST.
* @param p: You need find the successor node of p.
* @return: Successor of p.
*/
TreeNode * inorderSuccessor(TreeNode * root, TreeNode * p) {
if (root == NULL) return NULL;
TreeNode* fa = NULL;
while(root && root->val != p->val) {
if(root->val < p->val) root = root->right;
else if (root->val > p->val) {
fa = root;
root = root->left;
}
}
TreeNode* current = root;
if(root == NULL) return NULL;
if(root->right == NULL) return fa;
else {
root = root->right;
while(root->left) root= root->left;
return root;
}
}
};