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

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

  1. The number of nodes in the tree is between 

    1

     and 

    100

    .
  2. Each node will have value between   and 

    100

    .
  3. The given tree is a binary search tree.

Solution:

    修改二叉樹,使得每個節點的新值為原始樹中大于等于該節點值的和。

1.笨辦法,把每個都拿出來比,找到比它大的取出來。

/**
 * 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);
        }
    }
}
           

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