Path Sum II

描述

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,

  1.           5
  2.          / \
  3.         4   8
  4.        /   / \
  5.       11  13  4
  6.      /  \    / \
  7.     7    2  5   1

return

  1. [
  2.    [5,4,11,2],
  3.    [5,8,4,5]
  4. ]

分析

跟上一题相比，本题是求路径本身。且要求出所有结果，左子树求到了满意结果，不能return，要接着求右子树。

代码

  1. // Path Sum II
  2. // 时间复杂度O(n)，空间复杂度O(logn)
  3. public class Solution {
  4.     public List<List<Integer>> pathSum(TreeNode root, int sum) {
  5.         List<List<Integer>> result = new ArrayList<>();
  6.         ArrayList<Integer>  cur = new ArrayList<>(); // 中间结果
  7.         pathSum(root, sum, cur, result);
  8.         return result;
  9.     }
  10.     private static void pathSum(TreeNode root, int gap, ArrayList<Integer> cur,
  11.                                 List<List<Integer>> result) {
  12.         if (root == null) return;
  14.         cur.add(root.val);
  16.         if (root.left == null && root.right == null) { // leaf
  17.             if (gap == root.val)
  18.                 result.add(new ArrayList<>(cur));
  19.         }
  20.         pathSum(root.left, gap - root.val, cur, result);
  21.         pathSum(root.right, gap - root.val, cur, result);
  23.         cur.remove(cur.size() - 1);
  24.     }
  25. }

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/binary-tree/recursion/path-sum-ii.html