lintcode 访问路径
http://www.lintcode.com/zh-cn/problem/binary-tree-paths/
描述
给一棵二叉树，找出从根节点到叶子节点的所有路径。
样例
给出下面这棵二叉树：
所有根到叶子的路径为：
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| [
"1->2->5",
"1->3"
]
|
Java代码实现
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| /**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root the root of the binary tree
* @return all root-to-leaf paths
*/
public List<String> binaryTreePaths(TreeNode root) {
// Write your code here
List<String> result = new ArrayList<>();
if (root == null) {
return result;
}
String path = "";
paths(root, path, result);
return result;
}
public void paths(TreeNode root, String path, List<String> result) {
if (root == null) {
return;
}
if (root.left == null && root.right == null) {
if ("".equals(path)) {
path += root.val;
} else {
path += "->" + root.val;
}
result.add(path);
return;
}
if ("".equals(path)) {
path += root.val;
} else {
path += "->" + root.val;
}
paths(root.left, path, result);
paths(root.right, path, result);
}
}
|
Python代码实现
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| """
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
# @param {TreeNode} root the root of the binary tree
# @return {List[str]} all root-to-leaf paths
def binaryTreePaths(self, root):
# Write your code here
result = []
if root is None:
return result
path = ""
self.paths(root, path, result)
return result
def paths(self, root, path, result):
if root is None:
return
if root.left is None and root.right is None:
if path == "":
path = str(root.val)
else:
path = str(path) + "->" + str(root.val)
result.append(path)
return;
if path == "":
path = str(root.val)
else:
path = str(path) + "->" + str(root.val)
self.paths(root.left, path, result)
self.paths(root.right, path, result)
|