Path Sum I && II & III
Path Sum I
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22
,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) return false;
if (root.left == null && root.right == null && root.val == sum) return true;
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22
,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> all = new ArrayList<>();
helper(root, new ArrayList<Integer>(), all, , sum);
return all;
} private void helper(TreeNode root, List<Integer> list, List<List<Integer>> all, int currSum, int sum) {
// exit condition
if (root == null) return; list.add(root.val);
currSum += root.val; if (currSum == sum && root.left == null && root.right == null) {
all.add(new ArrayList<Integer>(list));
} helper(root.left, list, all, currSum, sum);
helper(root.right, list, all, currSum, sum);
list.remove(list.size() - );
}
}
Path Sum III
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
public class Solution {
public int pathSum(TreeNode root, int sum) {
int[] arr = new int[];
preOrder(root, sum, arr);
return arr[];
} public void preOrder(TreeNode root, int sum, int[] count) {
if (root == null) return;
printSums(root, sum, , count);
preOrder(root.left, sum, count);
preOrder(root.right, sum, count);
} public void printSums(TreeNode root, int sum, int currentSum, int[] count) {
if (root == null) return;
currentSum += root.val;
if (currentSum == sum) {
count[]++;
}
printSums(root.left, sum, currentSum, count);
printSums(root.right, sum, currentSum, count);
}
}
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