112. Path Sum

dici5041

于 2019-09-04 06:44:00 发布

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原文链接：http://www.cnblogs.com/wentiliangkaihua/p/11456721.html

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本文探讨了二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的数值。通过递归算法实现，判断条件为当前节点是否为空及左右子树是否存在符合条件的路径。

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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.

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.left == null && root.right == null) return sum == root.val;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}