112:路径总和

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huangge1199 2021-10-19 15:12:15 +08:00
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//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
//
// 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
//
//
// 示例 2
//
//
//输入root = [1,2,3], targetSum = 5
//输出false
//
//
// 示例 3
//
//
//输入root = [1,2], targetSum = 0
//输出false
//
//
//
//
// 提示
//
//
// 树中节点的数目在范围 [0, 5000]
// -1000 <= Node.val <= 1000
// -1000 <= targetSum <= 1000
//
// Related Topics 深度优先搜索 二叉树 👍 702 👎 0
package leetcode.editor.cn;
import com.code.leet.entiy.TreeNode;
//112:路径总和
class PathSum {
public static void main(String[] args) {
//测试代码
Solution solution = new PathSum().new Solution();
}
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
/**
* 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 dfs(root, 0, targetSum);
}
private boolean dfs(TreeNode root, int sum, int targetSum) {
if (root.left == null && root.right == null) {
return root.val + sum == targetSum;
}
if (root.left != null) {
if (dfs(root.left, sum + root.val, targetSum)) {
return true;
}
}
if (root.right != null) {
return dfs(root.right, sum + root.val, targetSum);
}
return false;
}
}
//leetcode submit region end(Prohibit modification and deletion)
}

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<p>给你二叉树的根节点 <code>root</code> 和一个表示目标和的整数 <code>targetSum</code> ,判断该树中是否存在 <strong>根节点到叶子节点</strong> 的路径,这条路径上所有节点值相加等于目标和 <code>targetSum</code></p>
<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>
<p> </p>
<p><strong>示例 1</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入:</strong>root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>输出:</strong>true
</pre>
<p><strong>示例 2</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" />
<pre>
<strong>输入:</strong>root = [1,2,3], targetSum = 5
<strong>输出:</strong>false
</pre>
<p><strong>示例 3</strong></p>
<pre>
<strong>输入:</strong>root = [1,2], targetSum = 0
<strong>输出:</strong>false
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li>树中节点的数目在范围 <code>[0, 5000]</code></li>
<li><code>-1000 <= Node.val <= 1000</code></li>
<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
<div><div>Related Topics</div><div><li></li><li>深度优先搜索</li><li>二叉树</li></div></div><br><div><li>👍 702</li><li>👎 0</li></div>