Given the root
of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.
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: root = [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]
思路:reverse inorder traverse,也就是inorder: left, current, right;
现在需要right, current, left, 然后value进行累加。逆着写inorder就可以了。
/**
* 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) {
Stack<TreeNode> stack = new Stack<>();
if(root == null) {
return null;
}
pushRight(root, stack);
TreeNode oldroot = root;
TreeNode pre = null;
while(!stack.isEmpty()) {
TreeNode node = stack.pop();
if(pre == null) {
pre = node;
} else {
node.val += pre.val;
pre = node;
}
//反着写inorder,就是push right,那么就是反着来,push left的right;
if(node.left != null) {
pushRight(node.left, stack);
}
}
return oldroot;
}
public void pushRight(TreeNode node, Stack<TreeNode> stack) {
while(node != null) {
stack.push(node);
node = node.right;
}
}
}