Recover Binary Search Tree [leetcode]

标签：leetcode

本题是在中序遍历的基础上，找不合规范（不是递增）的树节点对，然后交换

首先看两种序列：

1. 1 3 2 4=>应该交换3和2

2. 4 3 2 1=>应交换4和1

这两种序列对应了不符合条件的BST的中序遍历序列的所有可能性---两个节点中序遍历相邻/不相邻

如果我们用一个数组swap保存所有中序遍历不递增的结果，那么这个数组只可能是2或者4的大小

而我们交换数组中节点对内容，只需交换第一个和最后一个树节点对的内容

    vector<TreeNode *> swap;
    TreeNode * pre;
    void recoverTree(TreeNode *root) 
    {
        pre = new TreeNode(-999999);
        swap = vector<TreeNode *>();
        inorder(root);
        if (swap.size() > 1)
        {
            int val = swap[0]->val;
            swap[0]->val = swap[swap.size() - 1]->val;
            swap[swap.size() - 1]->val = val;
        }
    }
    
    void inorder(TreeNode * root)
    {
        if (root == NULL) return;
        inorder(root->left);
        if (pre->val > root->val)
        {
            swap.push_back(pre);
            swap.push_back(root);
        }
        pre = root;
        inorder(root->right);
    }

if(!first) first = pre;

last = root;

替换

swap.push_back(pre); swap.push_back(root);

代码如下：

TreeNode * first, * last;
    TreeNode * pre;
    void recoverTree(TreeNode *root) 
    {
        pre = new TreeNode(-999999);
        first = last = NULL;
        inorder(root);
        if (first)
        {
            int val = first->val;
            first->val = last->val;
            last->val = val;
        }
    }
    
    void inorder(TreeNode * root)
    {
        if (root == NULL) return;
        inorder(root->left);
        if (pre->val > root->val)
        {
            if (!first) first = pre;
            last = root;
        }
        pre = root;
        inorder(root->right);
    }

Recover Binary Search Tree [leetcode]

标签：leetcode

原文地址：http://blog.csdn.net/peerlessbloom/article/details/39611109