刷题顺序按照代码随想录建议
112.路径总和
题目描述
英文版描述
Given the root
of a binary tree and an integer targetSum
, return true
if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum
.
A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: The root-to-leaf path with the target sum is shown.
Example 2:
Input: root = [1,2,3], targetSum = 5 Output: false Explanation: There two root-to-leaf paths in the tree: (1 --> 2): The sum is 3. (1 --> 3): The sum is 4. There is no root-to-leaf path with sum = 5.
Example 3:
Input: root = [], targetSum = 0 Output: false Explanation: Since the tree is empty, there are no root-to-leaf paths.
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]
. -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
英文版地址
中文版描述
给你二叉树的根节点 root
和一个表示目标和的整数 targetSum
。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 targetSum
。如果存在,返回 true
;否则,返回 false
。
叶子节点 是指没有子节点的节点。
示例 1:
输入: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 输出: true 解释: 等于目标和的根节点到叶节点路径如上图所示。
示例 2:
输入: root = [1,2,3], targetSum = 5 输出: false 解释: 树中存在两条根节点到叶子节点的路径: (1 --> 2): 和为 3 (1 --> 3): 和为 4 不存在 sum = 5 的根节点到叶子节点的路径。
示例 3:
输入: root = [], targetSum = 0 输出: false 解释: 由于树是空的,所以不存在根节点到叶子节点的路径。
提示:
- 树中节点的数目在范围
[0, 5000]
内 -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
中文版地址
解题方法
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
if (root == null) {
return false;
}
return recursion(root, targetSum);
}
private boolean recursion(TreeNode root, int targetSum) {
if (targetSum- root.val == 0 && root.right == null && root.left == null) {
return true;
}
if (root.right == null && root.left == null) {
return false;a
}
if (root.left != null) {
if (recursion(root.left, targetSum - root.val)) {
return true;
}
}
if (root.right != null) {
if (recursion(root.right, targetSum - root.val)) {
return true;
}
}
return false;
}
}
复杂度分析
- 时间复杂度:O(n),其中 n 是二叉树的节点数,每一个节点恰好被遍历一次
- 空间复杂度:O(n),为递归过程中栈的开销,平均情况下为 O(logn),最坏情况下树呈现链状,为 O(n)
113.路径总和ii
题目描述
英文版描述
Given the root
of a binary tree and an integer targetSum
, return all root-to-leaf paths where the sum of the node values in the path equals targetSum
. Each path should be returned as a list of the node values , not node references.
A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 Output: [[5,4,11,2],[5,8,4,5]] Explanation: There are two paths whose sum equals targetSum: 5 + 4 + 11 + 2 = 22 5 + 8 + 4 + 5 = 22
Example 2:
Input: root = [1,2,3], targetSum = 5 Output: []
Example 3:
Input: root = [1,2], targetSum = 0 Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]
. -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
英文版地址
中文版描述
给你二叉树的根节点 root
和一个整数目标和 targetSum
,找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
叶子节点 是指没有子节点的节点。
示例 1:
输入: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 输出: [[5,4,11,2],[5,8,4,5]]
示例 2:
输入: root = [1,2,3], targetSum = 5 输出: []
示例 3:
输入: root = [1,2], targetSum = 0 输出: []
提示:
- 树中节点的数目在范围
[0, 5000]
内 -1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
中文版地址
解题方法
java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
List<List<Integer>> result = new ArrayList<>();
List<Integer> ll = new ArrayList<>();
/**
* 输入:root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
* 输出:[[5,4,11,2],[5,8,4,5]]
*
* @param root
* @param targetSum
* @return
*/
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
if (root == null) {
return result;
}
recursion(root, targetSum);
return result;
}
private void recursion(TreeNode root, int targetSum) {
ll.add(root.val);
if (root.left == null && root.right == null && targetSum-root.val == 0) {
result.add(new ArrayList<>(ll));
return;
}
if (root.left == null && root.right == null) {
return;
}
if (root.left != null) {
recursion(root.left, targetSum - root.val);
ll.remove(ll.size() - 1);
}
if (root.right != null) {
recursion(root.right, targetSum - root.val);
ll.remove(ll.size() - 1);
}
}
}
复杂度分析
- 时间复杂度:O(n),其中 n 是树的节点数。在最坏情况下,树的上半部分为链状,下半部分为完全二叉树,并且从根节点到每一个叶子节点的路径都符合题目要求。此时,路径的数目为 O(n),并且每一条路径的节点个数也为 O(n),因此要将这些路径全部添加进答案中,时间复杂度为 O(n^2)
- 空间复杂度:O(n),其中 n 是树的节点数,空间复杂度主要取决于栈空间的开销,栈中的元素个数不会超过树的节点数