538. Convert BST to Greater Tree
Tags: ‘Tree’
Given 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 sum of all keys greater than the original key in BST.
Example:
Input: The root of a Binary Search Tree like this:
5
/ \
2 13
Output: The root of a Greater Tree like this:
18
/ \
20 13
Solution
/// Reverse of Inorder Traversal Recursive 99% 37%
class Solution {
private int sum = 0;
public TreeNode convertBST(TreeNode root) {
if (root == null) return null;
convertBST(root.right);
root.val += sum;
sum = root.val;
convertBST(root.left);
return root;
}
}
// Reverse of Inorder Traversal 14% 37%
class Solution {
public TreeNode convertBST(TreeNode root) {
TreeNode curr = root;
Stack<TreeNode> stack = new Stack<>();
int sum = 0;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
stack.push(curr);
curr = curr.right;
}
curr = stack.pop();
sum += curr.val;
curr.val = sum;
curr = curr.left;
}
return root;
}
}
//Two Pass, 6% 6%
class Solution {
public TreeNode convertBST(TreeNode root) {
// First pass calculate sum
// Preorder Traversal
int sum = 0;
Stack<TreeNode> stack = new Stack<>();
TreeNode curr = root;
stack.push(curr);
while (!stack.isEmpty()) {
curr = stack.pop();
if (curr != null) {
sum += curr.val;
stack.push(curr.right);
stack.push(curr.left);
}
}
// Second psss, rebuild tree value
// Inorder traversal, calculate currentSum(includeing current node)
// for each node, newValue = sum - currentSum
int currentSum = 0;
curr = root;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
curr = stack.pop();
currentSum += curr.val;
curr.val = sum - currentSum + curr.val;
curr = curr.right;
}
return root;
}
}