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算法笔记: 力扣#538 把二叉搜索树转换为累加树

admin 11月 12, 2020 0

问题描述

解法

分析

Python 实现

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def __init__(self):
        self._sum = 0
        
    def convertBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if root is None:
            return root
        self.convertBST(root.right)
        self._sum += root.val
        root.val = self._sum
        self.convertBST(root.left)
        return root

Java 实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    private int sum = 0;
    
    public TreeNode convertBST(TreeNode root) {
        if(root == null){ return root; }
        convertBST(root.right);
        this.sum += root.val;
        root.val = this.sum;
        convertBST(root.left);
        return root;
    }
}

时间复杂度

O(n).

空间复杂度

O(n).

链接

538. Convert BST to Greater Tree
538. 把二叉搜索树转换为累加树
 (English version) Algorithm Notes: Leetcode#538 Convert BST to Greater Tree