lintcode 访问路径
http://www.lintcode.com/zh-cn/problem/binary-tree-path-sum/
描述
给定一个二叉树，找出所有路径中各节点相加总和等于给定目标值的路径。一个有效的路径，指的是从根节点到叶节点的路径。
样例
给定一个二叉树，和目标值 = 5：
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5
| 1
/ \
2 4
/ \
2 3
|
返回：
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| [
[1, 2, 2],
[1, 4]
]
|
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 binary tree
* @param target an integer
* @return all valid paths
*/
public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
// Write your code here
List<List<Integer>> result = new ArrayList<List<Integer>>();
List<Integer> path = new ArrayList<Integer>();
search(root, target, path, result);
return result;
}
private void search(TreeNode root, int target, List<Integer> path, List<List<Integer>> result) {
if (root == null) {
return;
}
if (root.left == null && root.right == null && root.val == target) {
path.add(root.val);
result.add(path);
return;
}
if (root.left != null) {
List<Integer> left = new ArrayList<Integer>(path);
left.add(root.val);
search(root.left, target - root.val, left, result);
}
if (root.right != null) {
List<Integer> right = new ArrayList<Integer>(path);
right.add(root.val);
search(root.right, target - root.val, right, result);
}
}
}
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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 binary tree
# @param {int} target an integer
# @return {int[][]} all valid paths
def binaryTreePathSum(self, root, target):
# Write your code here
result = []
path = []
self.search(root, target, path, result)
return result
def search(self, root, target, path, result):
if root is None:
return
if root.left is None and root.right is None and root.val == target:
path.append(root.val)
result.append(path)
return
if root.left:
left = path[:]
left.append(root.val)
self.search(root.left, target - root.val, left, result)
if root.right:
right = path[:]
right.append(root.val)
self.search(root.right, target - root.val, right, result)
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