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面试题 34. 二叉树中和为某一值的路径

题目描述

输入一棵二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。

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]
]

提示：

1. 节点总数 <= 10000

解法

先序遍历+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);
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} sum
 * @return {number[][]}
 */
var pathSum = function (root, sum) {
  if (!root) return [];
  let res = [];
  function dfs(node, sum, arr) {
    if (!node) return;
    arr = [...arr, node.val];
    if (node.val === sum && !node.left && !node.right) {
      res.push(arr);
      return;
    }
    dfs(node.left, sum - node.val, arr);
    dfs(node.right, sum - node.val, arr);
  }
  dfs(root, sum, []);
  return res;
};

Go

var res [][]int
func pathSum(root *TreeNode, sum int) [][]int {
    res = [][]int{}
    if root == nil {
        return res
    }
    helper(root, sum, []int{})
    return res
}

func helper(node *TreeNode, target int, ans []int) {
    if node == nil {
        return
    }
    ans = append(ans,node.Val)
    target -= node.Val
    if target == 0 && node.Left == nil && node.Right == nil {
        tmp := make([]int,len(ans))
        copy(tmp,ans)
        res = append(res,tmp)
    } else {
        helper(node.Left, target, ans)
        helper(node.Right, target, ans)
    }
}

C++

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int target) {
        vector<vector<int>> ans;
        vector<int> path;
        dfs(root, ans, path, target);
        return ans;
    }

    void dfs(TreeNode* root, vector<vector<int>>& ans, vector<int>& path, int target) {
        if (root == NULL) {
            return;
        }
        target -= root->val;
        path.push_back(root->val);
        if (root->left == NULL && root->right == NULL) {
            if (target == 0) {
                ans.push_back(vector<int>(path));
            }
        }
        dfs(root->left, ans, path, target);
        dfs(root->right, ans, path, target);
        path.pop_back();
    }
};

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