LeetCode the binary search tree into a tree cumulative

The first 538 title

给定一个二叉搜索树（Binary Search Tree），把它转换成为累加树（Greater Tree)，使得每个节点的值是原来的节点值加上所有大于它的节点值之和。

例如：

输入: 二叉搜索树:
              5
            /   \
           2     13

输出: 转换为累加树:
             18
            /   \
          20     13

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/convert-bst-to-greater-tree

concept

Binary search tree

二叉查找树（Binary Search Tree），（又：二叉搜索树，二叉排序树）它或者是一棵空树，或者是具有下列性质的二叉树： 若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值； 若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值； 它的左、右子树也分别为二叉排序树。

Preorder

中序遍历（LDR）是二叉树遍历的一种，也叫做中根遍历、中序周游。在二叉树中，中序遍历首先遍历左子树，然后访问根结点，最后遍历右子树。

Problem-solving ideas

• By definition we know that, in order to traverse a BST increasing sequence
• We will preorder inverted (first left sub-tree, the root node, then the right subtree), you can achieve a descending sequence
• The traversal sequence, and to update the accumulated small node values ​​from a large, accumulated to achieve tree (original node = node value plus all values ​​greater than the sum of its node value)

Code

1. The idea of ​​the recursive traversal sequence inverted achieved

//递归实现
class Solution1 {
    public TreeNode convertBST(TreeNode root) {
        addSum(root, 0);
        return root;
    }

    public int addSum(TreeNode node, int parentVal) {
        //如果没有节点了，返回父节点值
        if (node == null) {
            return parentVal;
        }

        //累加右边所有节点值
        int rVal = addSum(node.right, parentVal);

        //当前节点值=右边所有节点累加值+当前节点值
        node.val += rVal;
        //System.out.println("当前节点值：" + node.val);

        //累加左边所有节点值
        int lVal = addSum(node.left, node.val);
        return lVal;
    }
}

2. Use the stack, to recursive Implementation

//利用堆栈，去递归化
class Solution2 {
    public TreeNode convertBST(TreeNode root) {
        TreeNode oRoot = root;
        Stack<TreeNode> stack = new Stack();
        int sum = 0;
        while (true) {
            //右节点入栈
            while (root != null) {
                stack.push(root);
                root = root.right;
            }

            //如果栈为空退出循环
            if (stack.empty()) {
                break;
            }
            //否则出栈进入计算
            else {
                TreeNode node = stack.pop();
                //更新节点值
                node.val += sum;
                //更新sum值
                sum = node.val;
                //左节点进入[右节点入栈]
                root = node.left;
            }
        }
        //返回原树，此时该树所有节点已做更新
        return oRoot;
    }
}

to sum up

We achieve by submitting the code found in the stack will be much slower than the recursive execution efficiency, because:

• Jvm tail recursion is optimized compiler recognizes treated and for which the corresponding iteration
• Stack implementation requires frequent push (stack), pop (pop) operation causes performance degradation

data

• Example Source
• Original Address
• Spring Boot, Spring Cloud instance learning