题目
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] ]分析
这题直接用递归写,类似DFS。
代码
import java.util.ArrayList;
public class PathSumII {
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
private ArrayList<ArrayList<Integer>> results;
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
results = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> list = new ArrayList<Integer>();
solve(root, sum, list);
return results;
}
private void solve(TreeNode root, int sum, ArrayList<Integer> list) {
if (root == null) {
return;
}
list.add(root.val);
if (sum == root.val && root.left == null && root.right == null) {
results.add(new ArrayList<Integer>(list));
} else {
solve(root.left, sum - root.val, list);
solve(root.right, sum - root.val, list);
}
list.remove(list.size() - 1);
}
}
本文介绍了一种算法,用于查找二叉树中所有从根节点到叶子节点的路径,这些路径上的数值之和等于给定的目标值。通过递归方式实现,类似于深度优先搜索。
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