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
深度优先搜索+路径记录。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def binaryTreePaths(self, root: TreeNode) -> List[str]:
def dfs(root):
if root is None:
return
path.append(str(root.val))
if root.left is None and root.right is None:
res.append("->".join(path))
dfs(root.left)
dfs(root.right)
path.pop()
res = []
path = []
dfs(root)
return res
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private List<String> res;
private List<String> path;
public List<String> binaryTreePaths(TreeNode root) {
if (root == null) return Collections.emptyList();
res = new ArrayList<>();
path = new ArrayList<>();
dfs(root);
return res;
}
private void dfs(TreeNode root) {
if (root == null) return;
path.add(String.valueOf(root.val));
if (root.left == null && root.right == null) {
res.add(String.join("->", path));
}
dfs(root.left);
dfs(root.right);
path.remove(path.size() - 1);
}
}
/**
* 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 pre = '';
dfs(root, pre, ans);
return ans;
};
function dfs(root: TreeNode | null, pre: string, ans: string[]): void {
if (root == null) return;
let val = String(root.val);
pre = pre.length > 0 ? `${pre}->${val}` : pre + val;
// 叶子节点
if (root.left == null && root.right == null) {
ans.push(pre);
return;
}
dfs(root.left, pre, ans);
dfs(root.right, pre, ans);
}