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113. 路径总和 II

English Version

题目描述

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:
给定如下二叉树，以及目标和 sum = 22，

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

返回:

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

解法

深度优先搜索+路径记录。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        def dfs(root, sum):
            if root is None:
                return
            path.append(root.val)
            if root.val == sum and root.left is None and root.right is None:
                res.append(path.copy())
            dfs(root.left, sum - root.val)
            dfs(root.right, sum - root.val)
            path.pop()
        if not root:
            return []
        res = []
        path = []
        dfs(root, sum)
        return res

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private List<List<Integer>> res;
    private List<Integer> path;

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        if (root == null) return Collections.emptyList();
        res = new ArrayList<>();
        path = new ArrayList<>();
        dfs(root, sum);
        return res;
    }

    private void dfs(TreeNode root, int sum) {
        if (root == null) return;
        path.add(root.val);
        if (root.val == sum && root.left == null && root.right == null) {
            res.add(new ArrayList<>(path));
        }
        dfs(root.left, sum - root.val);
        dfs(root.right, sum - root.val);
        path.remove(path.size() - 1);
    }
}

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