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  • 0538. Convert BST to Greater Tree (M)

    Convert BST to Greater Tree (M)

    题目

    Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

    As a reminder, a binary search tree is a tree that satisfies these constraints:

    • The left subtree of a node contains only nodes with keys less than the node's key.
    • The right subtree of a node contains only nodes with keys greater than the node's key.
    • Both the left and right subtrees must also be binary search trees.

    Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

    Example 1:

    Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
    Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
    

    Example 2:

    Input: root = [0,null,1]
    Output: [1,null,1]
    

    Example 3:

    Input: root = [1,0,2]
    Output: [3,3,2]
    

    Example 4:

    Input: root = [3,2,4,1]
    Output: [7,9,4,10]
    

    Constraints:

    • The number of nodes in the tree is in the range [0, 10^4].
    • -10^4 <= Node.val <= 10^4
    • All the values in the tree are unique.
    • root is guaranteed to be a valid binary search tree.

    题意

    对一个二叉搜索树T中的结点进行修改,使每个结点的值变为原值x加上T中所有大于x的值的和。

    思路

    由于BST的中序遍历序列S是递增,每个结点的值实际上变成了原值x加上S中所有排在x后面的值,因此可以从S中最后一个结点开始向前处理,逐步累加值即可,做一次"逆"中序遍历。


    代码实现

    Java

    class Solution {
        private int sum = 0;
    
        public TreeNode convertBST(TreeNode root) {
            if (root == null) return null;
    
            convertBST(root.right);
            sum += root.val;
            root.val = sum;
            convertBST(root.left);
    
            return root;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/14393430.html
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