112 Path Sum 路径总和
Description:
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
题目描述:
给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，这条路径上所有节点值相加等于目标和。
说明: 叶子节点是指没有子节点的节点。
示例：
给定如下二叉树，以及目标和 sum = 22，
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
思路:
参考LeetCode #104 Maximum Depth of Binary Tree 二叉树的最大深度
注意下递归结束的条件
- 递归
- 迭代
时间复杂度O(n), 空间复杂度O(n), n为树中结点数
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
bool hasPathSum(TreeNode* root, int sum)
{
if (!root) return false;
if (!root -> left and !root -> right) return root -> val == sum;
return hasPathSum(root -> left, sum - root -> val) or hasPathSum(root -> right, sum - root -> val);
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) return false;
if (root.left == null && root.right == null) return root.val == sum;
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if not root:
return False
stack = [(root, sum)]
while stack:
cur, temp = stack.pop()
if not cur.left and not cur.right and cur.val == temp:
return True
if cur.left:
stack.append([cur.left, temp - cur.val])
if cur.right:
stack.append([cur.right, temp - cur.val])
return False
