Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum 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 5 1
Return:
[
[5,4,11,2],
[5,8,4,5]
]
难度：medium
题目：给定二叉树与一个数和 sum，找出所有由根到叶结点的和为 sum 的路径
思路：递归（前序遍历）
Runtime: 2 ms, faster than 56.03% of Java online submissions for Path Sum II.Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Path Sum II.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {val = x;}
* }
*/
class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new ArrayList<>();
pathSum(root, sum, new Stack<>(), result);
return result;
}
private void pathSum(TreeNode root, int sum, Stack<Integer> stack, List<List<Integer>> result) {
if (null != root) {
if (null == root.left && null == root.right) {
if (sum == root.val) {
stack.push(root.val);
result.add(new ArrayList<>(stack));
stack.pop();
}
} else {
stack.push(root.val);
pathSum(root.left, sum – root.val, stack, result);
pathSum(root.right, sum – root.val, stack, result);
stack.pop();
}
}
}
}