112. Path Sum - Easy

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.

time: O(n), space: O(height)

/**
 * 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 sum - root.val == 0;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

原文地址：https://www.cnblogs.com/fatttcat/p/10198680.html