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>> res = new ArrayList<List<Integer>>();
List<Integer> list = new ArrayList<Integer>();
solve(root, sum, res, list);
return res;
}
private void solve(TreeNode root, int sum, List<List<Integer>> res,
List<Integer> list) {
if (root == null) {
return;
}
list.add(root.val);
sum -= root.val;
if (root.left == null && root.right == null) {
if (sum == 0) {
res.add(new ArrayList<Integer>(list));
}
} else {
if (root.left != null) {
solve(root.left, sum, res, list);
}
if (root.right != null) {
solve(root.right, sum, res, list);
}
}
list.remove(list.size()-1);
}
}
本文介绍了一种算法,用于在给定的二叉树中找到所有从根节点到叶子节点的路径,使得这些路径上的节点值之和等于给定的数值。通过递归地遍历树结构,算法能够有效地识别满足条件的路径。
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