Given a binary tree, return all root-to-leaf paths.
For example, given the following binary tree:
1
/ \
2 3
\
5
All root-to-leaf paths are:
["1->2->5", "1->3"]
给一个二叉树，返回所有根到叶节点的路径。
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<String> binaryTreePaths(TreeNode root) {
List<String> list = new ArrayList<>();
binaryTreePathsHelper(root, list, new String());
return list;
} public void binaryTreePathsHelper(TreeNode root, List<String> list, String string) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
string = string + root.val;
list.add(string);
return;
} binaryTreePathsHelper(root.left, list, string + root.val + "->");
binaryTreePathsHelper(root.right, list, string + root.val + "->");
}
}
Python:
class Solution:
# @param {TreeNode} root
# @return {string[]}
def binaryTreePaths(self, root):
result, path = [], []
self.binaryTreePathsRecu(root, path, result)
return result def binaryTreePathsRecu(self, node, path, result):
if node is None:
return if node.left is node.right is None:
ans = ""
for n in path:
ans += str(n.val) + "->"
result.append(ans + str(node.val)) if node.left:
path.append(node)
self.binaryTreePathsRecu(node.left, path, result)
path.pop() if node.right:
path.append(node)
self.binaryTreePathsRecu(node.right, path, result)
path.pop()
C++: DFS
class Solution {
public:
vector<string> binaryTreePaths(TreeNode* root) {
vector<string> res;
if (root) dfs(root, "", res);
return res;
}
void dfs(TreeNode *root, string out, vector<string> &res) {
out += to_string(root->val);
if (!root->left && !root->right) res.push_back(out);
else {
if (root->left) dfs(root->left, out + "->", res);
if (root->right) dfs(root->right, out + "->", res);
}
}
};
C++:
class Solution {
public:
vector<string> binaryTreePaths(TreeNode* root) {
if (!root) return {};
if (!root->left && !root->right) return {to_string(root->val)};
vector<string> left = binaryTreePaths(root->left);
vector<string> right = binaryTreePaths(root->right);
left.insert(left.end(), right.begin(), right.end());
for (auto &a : left) {
a = to_string(root->val) + "->" + a;
}
return left;
}
};
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[LeetCode] 113. Path Sum II 路径和 II
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