141.Path Sum

题目：

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.

给定二叉树和一个和，确定树是否具有根到叶路径，使得沿路径的所有值相加等于给定的总和。

Note: A leaf is a node with no children.

注意：叶子是没有子节点的节点。

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.

解答：

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

详解：

根—>叶子节点：DFS深度优先遍历