Cinte's Leetcode Record
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538. 把二叉搜索树转换为累加树

Posted on Edited on In Disqus:
Symbols count in article: 1.5k Reading time ≈ 1 mins.

538. 把二叉搜索树转换为累加树

中序遍历变种
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        int rightVal = 0;
        dfs(root, rightVal);
        return root;
    }
private:
    void dfs(TreeNode* root, int& rightVal)
    {
        if(!root)
            return;
        if(root->right)
        {
            dfs(root->right, rightVal);
        }
        root->val += rightVal;
        rightVal = root->val;
        if(root->left)
        {
            dfs(root->left, rightVal);
        }
    }
};

右->中->左

rightVal始终保存他右边树枝的最大值

我好傻

简单的思考方法:

中序单增，于是反中序为单减，对于每个节点加上他左边的和就行

迭代

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        stack<TreeNode*> q;
        int curSum = 0;
        TreeNode* ret = root;
        while(!q.empty() || root)
        {
            while(root)
            {
                q.push(root);
                root = root->right;
            }
            root = q.top();
            q.pop();
            curSum += root->val;
            root->val = curSum;
            root = root->left;
        }
        return ret;
    }
};