113. Path Sum II

摘要： 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 childre...

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
/ \
48
// \
11134
 /\/ \
7251

Return:

[
[5,4,11,2],
[5,8,4,5]
]

難度：medium

題目：給定二叉樹與一個數和sum，找出所有由根到葉結點的和為sum的路徑

思路：遞迴（前序遍歷）

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();
}
}
}
}