【改造中序遍历算法】1038. 从二叉搜索树到更大和树

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);}
}