Path Sum
Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
A leaf is a node with no children.
public boolean hasPathSum(TreeNode root, int sum) {
// iteration method
if (root == null) {return false;}
Stack<TreeNode> path = new Stack<>();
Stack<Integer> sub = new Stack<>();
path.push(root);
sub.push(root.val);
while (!path.isEmpty()) {
TreeNode temp = path.pop();
int tempVal = sub.pop();
if (temp.left == null && temp.right == null) {if (tempVal == sum) return true;}
else {
if (temp.left != null) {
path.push(temp.left);
sub.push(temp.left.val + tempVal);
}
if (temp.right != null) {
path.push(temp.right);
sub.push(temp.right.val + tempVal);
}
}
}
return false;
}Related Problems
You are given the heads of two sorted linked lists list1 and list2.
Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.
Return the head of the merged linked list.
Given the root of a binary tree, flatten the tree into a "linked list":
The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
The "linked list" should be in the same order as a pre-order traversal of the binary tree.
Given an m x n board of characters and a list of strings words, return all words on the board.
Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.
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.