動手刷LeetCode-恢復二叉搜索樹

即按照中序遍歷滿足 升序

知識點

對於中序遍歷：1、遞歸 2、遍歷 3、Morris
掌握前兩種方法實現 樹的遍歷；

解題思路

迭代

python實現:

class Solution:
    def recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        firstNode = None
        secondNode = None
        pre = TreeNode(float("-inf"))

        stack = []
        p = root
        while p or stack:
            while p:
                stack.append(p)
                p = p.left
            p = stack.pop()
            
            if not firstNode and pre.val > p.val:
                    firstNode = pre
            if firstNode and pre.val > p.val:
                #print(firstNode.val,pre.val, p.val)
                secondNode = p
            pre = p
            p = p.right
        firstNode.val, secondNode.val = secondNode.val, firstNode.val

C++ 實現：

class Solution {
public:
    void recoverTree(TreeNode* root) {
        // 使用迭代，就是利用棧的思想
        TreeNode* first =NULL;
        TreeNode* second=NULL;
        TreeNode* pre =new TreeNode(-1);// 最小值

        stack<TreeNode*> s ={}; // creat a empty stack

        TreeNode* p=root;// define a pointer to point the tree root

        while(p!=NULL || !s.empty()){
            while(p!=NULL){
                    s.push(p);
                    p=p->left;
            }
            p=s.top();// return the ding value
            s.pop();// delete it
            
            if(first==NULL and pre->val>p->val){
                first=pre;
            }
            if(first!=NULL and pre->val>p->val){
                second=p;
            }
            pre=p;
            p=p->right;
        }

        first->val=second->val;
        first->val,second->val=second->val,first->val;// i dont know what it means

    }
};

小結

管它什麼真理無窮，進一寸有一寸的歡喜；