【樹】B037_LC_從二叉搜索樹到更大和樹（反中序遍歷）

一、Problem

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.

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.

二、Solution

方法一：dfs

如果要你求所有小於當前結點的值的和等於當前結點的新值，那麼你很容易想到中序遍歷，反過來求這道題也是一樣的：

• 右路突到盡頭，然後返回，讓當前結點加上右子樹的總和，並記錄當前結點的總和 pre
• 左路突破，沒到達一個結點就加上 pre，繼續向左深入

class Solution {
	int pre;
	void dfs(TreeNode root) {
		if (root == null)
			return;
		dfs(root.right);
		root.val += pre;
		pre = root.val;
		dfs(root.left);
	}
    public TreeNode bstToGst(TreeNode root) {
    	dfs(root);
    	return root;
    }
}

複雜度分析

• 時間複雜度：$O(n)$，
• 空間複雜度：$O(1)$，