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257. 二叉树的所有路径

English Version

题目描述

给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

解法

深度优先搜索+路径记录。

Python3

# 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: TreeNode) -> List[str]:
        def dfs(root):
            if root is None:
                return
            t.append(str(root.val))
            if root.left is None and root.right is None:
                ans.append('->'.join(t))
            dfs(root.left)
            dfs(root.right)
            t.pop()

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

Java

/**
 * 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;
    private List<String> t;

    public List<String> binaryTreePaths(TreeNode root) {
        ans = new ArrayList<>();
        t = new ArrayList<>();
        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));
        }
        dfs(root.left);
        dfs(root.right);
        t.remove(t.size() - 1);
    }
}

TypeScript

/**
 * 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[] {
    let ans = [];
    let t = [];
    function dfs(root) {
        if (!root) return;
        t.push(String(root.val));
        if (!root.left && !root.right) ans.push(t.join('->'));
        dfs(root.left);
        dfs(root.right);
        t.pop();
    }
    dfs(root);
    return ans;
}

Go

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

C++

/**
 * 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> ans;

    vector<string> binaryTreePaths(TreeNode* root) {
        dfs(root, "");
        return ans;
    }

    void dfs(TreeNode* root, string t) {
        t += to_string(root->val);
        if (!root->left && !root->right)
        {
            ans.push_back(t);
            return;
        }
        t += "->";
        if (root->left) dfs(root->left, t);
        if (root->right) dfs(root->right, t);
    }
};

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