1038. 从二叉搜索树到更大和树
解题思路
- 改造中序遍历算法
- 先遍历右子树 然后累加当前节点的值 再遍历左子树
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode bstToGst(TreeNode root) {
// 改造中序遍历算法
traverse(root);
return root;
}
int sum = 0;
public void traverse(TreeNode root){
if(root == null){
return;
}
traverse(root.right);
sum += root.val;
root.val = sum;// 比他大的所有数字之和
traverse(root.left);
}
}