[leetcode]1038. Binary Search Tree to Greater Sum Tree

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本文介绍了一种算法，将二叉搜索树每个节点的值修改为原始树中所有大于或等于该节点值的节点值之和。通过先遍历获取所有节点值，排序后再次遍历树来更新节点值，确保了最终树的正确性。

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]

Note:

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.

Solution:

修改二叉树，使得每个节点的新值为原始树中大于等于该节点值的和。

1.笨办法，把每个都拿出来比，找到比它大的取出来。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> list = new ArrayList<>();
    public TreeNode bstToGst(TreeNode root) {
        if(root == null) {
            return null;
        }
        travel(root);
        Collections.sort(list);
        return travel2(root);
    }
    
    public TreeNode travel2(TreeNode root) {
        if(root != null) {
            root.val = calculate(list, root);
        }
        if(root.left != null) {
            travel2(root.left);
        }
        if(root.right != null) {
            travel2(root.right);
        }
        return root;
        
    }
    
    
     public int calculate(List<Integer> list, TreeNode root) {
        int len = list.size();
        int stop=0;
        int res=0;
        int i = 0;
        for(i = 0; i<len; i++) {
            if(root.val == list.get(i)) {
               break;
            }
            
        }
        stop=i;
        for(int j = stop; j < len; j++) {
            res += list.get(j);
        }
        return res;
    }
    
    public void travel(TreeNode root) {
        if(root != null) {
            list.add(root.val);
        }
        if(root.left != null) {
            travel(root.left);
        }
        if(root.right != null) {
            travel(root.right);
        }
    }
}