LeetCode: Recover Binary Search Tree [099]

发布时间：2020-12-14 04:34:43 所属栏目：大数据来源：网络整理

导读：【题目】 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? confus

【题目】

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.

【题意】

给定的二叉搜索树中有两个节点的值错换了，找出这两个节点。恢复二叉搜索树。要求不适用额外的空间。

【思路】

中序遍历二叉树。一棵正常的二叉树中序遍历得到有序的序列，现有两个节点的值的调换了，则肯定是一个较大的值被放到了序列的前段。而较小的值被放到了序列的后段。节点的错换使得序列中出现了s[i-1]>s[i]的情况。假设错换的点正好是相邻的两个数，则s[i-1]>s[i]的情况仅仅出现一次。假设不相邻，则会出现两次，第一次出现是前者为错换的较大值节点，第二次出现时后者为错换的较小值节点。

【代码】

/**
 * 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;}
        }
        //替换两个节点的值
        int temp=nodeLarge->val;
        nodeLarge->val = nodeSmall->val;
        nodeSmall->val = temp;
    }
};

（编辑：李大同）

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