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LeetCode: Recover Binary Search Tree [099]

阿新 • • 發佈：2019-04-06
uri blog 第一次 str article ack ear ria com

【題目】

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

confused what?"{1,#,2,3}"?means?

?> read more on how binary tree is serialized on OJ.



【題意】

給定的二叉搜索樹中有兩個節點的值錯換了，找出這兩個節點。恢復二叉搜索樹。要求不適用額外的空間。


【思路】

中序遍歷二叉樹。一棵正常的二叉樹中序遍歷得到有序的序列，現有兩個節點的值的調換了，則肯定是一個較大的值被放到了序列的前段。而較小的值被放到了序列的後段。節點的錯換使得序列中出現了s[i-1]>s[i]的情況。假設錯換的點正好是相鄰的兩個數，則s[i-1]>s[i]的情況僅僅出現一次。假設不相鄰，則會出現兩次，第一次出現是前者為錯換的較大值節點，第二次出現時後者為錯換的較小值節點。


【代碼】

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode *root) {
        stack<TreeNode*> st;
        TreeNode* pointer=root;
        TreeNode* prev=NULL;
        TreeNode* nodeLarge=NULL;
        TreeNode* nodeSmall=NULL;
        while(pointer){st.push(pointer); pointer=pointer->left;}
        while(!st.empty()){
            TreeNode* cur = st.top();
            st.pop();
            if(prev && prev->val > cur->val){
                if(nodeLarge==NULL || prev->val > nodeLarge->val) nodeLarge=prev;
                if(nodeSmall==NULL || cur->val < nodeSmall->val) nodeSmall=cur;
            }
            prev=cur;
            pointer=cur->right;
            while(pointer){st.push(pointer); pointer=pointer->left;}
        }
        //替換兩個節點的值
        int temp=nodeLarge->val;
        nodeLarge->val = nodeSmall->val;
        nodeSmall->val = temp;
    }
};


LeetCode: Recover Binary Search Tree [099]

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