标签:TreeNode val Sum leaf targetSum Path null root LeetCode
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.
Example1:
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.
Example2:
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.
Example3:
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
这道题的思路一看就是递归,但是需要注意的递归的出口,不但要处理null的情况,更要处理叶子结点。targetSum每次都会减掉当时的层结点的值,当到叶子的时候,只要叶子的值和剩下的targetSum一致,就可以了。其实就是前序遍历,每次把当前结点的val减掉。
注意:
- 这道题必须是到叶子结点累加,所以不需要考虑如果中间过程中有累加与targetSum相等。
/**
* 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;
}
targetSum -= root.val;
if (root.left == null && root.right == null) {
return targetSum == 0;
}
return hasPathSum(root.left, targetSum) || hasPathSum(root.right, targetSum);
}
}
标签:TreeNode,val,Sum,leaf,targetSum,Path,null,root,LeetCode 来源: https://www.cnblogs.com/codingEskimo/p/15780174.html
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