题目
Given the root of a binary search tree with distinct values, modify it so that every node has a new value equal to the sum of the values of the original tree that are greater than or equal to node.val.
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.
Example 1:
Input: [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]
Constraints:
The number of nodes in the tree is between 1 and 100.
Each node will have value between 0 and 100.
The given tree is a binary search tree.
分析
题意:从右到左、下到上累加和,赋给当前节点。
观察后可知,遍历顺序为“右、根、左”,因此,以这个为一个“单元”,设计递归算法即可。
解答
class Solution {
// 记录当前求和值
int sum=0;
public TreeNode bstToGst(TreeNode root) {
if(root==null) return null;
// 遍历右
bstToGst(root.right);
// 遍历根
sum+=root.val;
root.val=sum;
// 遍历左
bstToGst(root.left);
return root;
}
}
评论区的答案跟我的一样。
递归小结
这类递归问题,找出一个符合条件的最小递归单元进行调试即可。
理解递归其实很简单,高中的时候都学过数学归纳法,我当初理解递归就是类比数学归纳法。
数学归纳法的常见套路就是
1.当n=1时,显然成立.
2.假设当n=k时(把式中n换成k,写出来)成立, 则当n=k+1时, 该式也成立。
按我的理解,上面这就是一个“递推最小单元”,也就是“递归最小单元”。只要上面的最小单元成立,那么递归无数层,都是成立的。
写递归算法,就得找出一个最小单元,对其写一个算法,然后进行递归。(区别就在于,设计算法的时候需要把递归的步骤以及特征考虑进去。)