Question

257. 二叉树的所有路径

English Version

题目描述

给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。

叶子节点 是指没有子节点的节点。

 

示例 1：

输入：root = [1,2,3,null,5]
输出：["1->2->5","1->3"]

示例 2：

输入：root = [1]
输出：["1"]

 

提示：

• 树中节点的数目在范围 [1, 100] 内
• -100 <= Node.val <= 100

解法

方法一：DFS

我们可以使用深度优先搜索的方法遍历整棵二叉树，每一次我们将当前的节点添加到路径中。如果当前的节点是叶子节点，则我们将整个路径加入到答案中。否则我们继续递归遍历节点的孩子节点。最后当我们递归结束返回到当前节点时，我们需要将当前节点从路径中删除。

时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
    def dfs(root: Optional[TreeNode]):
      if root is None:
        return
      t.append(str(root.val))
      if root.left is None and root.right is None:
        ans.append("->".join(t))
      else:
        dfs(root.left)
        dfs(root.right)
      t.pop()

    ans = []
    t = []
    dfs(root)
    return ans

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private List<String> ans = new ArrayList<>();
  private List<String> t = new ArrayList<>();

  public List<String> binaryTreePaths(TreeNode root) {
    dfs(root);
    return ans;
  }

  private void dfs(TreeNode root) {
    if (root == null) {
      return;
    }
    t.add(root.val + "");
    if (root.left == null && root.right == null) {
      ans.add(String.join("->", t));
    } else {
      dfs(root.left);
      dfs(root.right);
    }
    t.remove(t.size() - 1);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  vector<string> binaryTreePaths(TreeNode* root) {
    vector<string> ans;
    vector<string> t;
    function<void(TreeNode*)> dfs = [&](TreeNode* root) {
      if (!root) {
        return;
      }
      t.push_back(to_string(root->val));
      if (!root->left && !root->right) {
        ans.push_back(join(t));
      } else {
        dfs(root->left);
        dfs(root->right);
      }
      t.pop_back();
    };
    dfs(root);
    return ans;
  }

  string join(vector<string>& t, string sep = "->") {
    string ans;
    for (int i = 0; i < t.size(); ++i) {
      if (i > 0) {
        ans += sep;
      }
      ans += t[i];
    }
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func binaryTreePaths(root *TreeNode) (ans []string) {
  t := []string{}
  var dfs func(*TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    t = append(t, strconv.Itoa(root.Val))
    if root.Left == nil && root.Right == nil {
      ans = append(ans, strings.Join(t, "->"))
    } else {
      dfs(root.Left)
      dfs(root.Right)
    }
    t = t[:len(t)-1]
  }
  dfs(root)
  return
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function binaryTreePaths(root: TreeNode | null): string[] {
  const ans: string[] = [];
  const t: number[] = [];
  const dfs = (root: TreeNode | null) => {
    if (!root) {
      return;
    }
    t.push(root.val);
    if (!root.left && !root.right) {
      ans.push(t.join('->'));
    } else {
      dfs(root.left);
      dfs(root.right);
    }
    t.pop();
  };
  dfs(root);
  return ans;
}