[LeetCode] 112. Path Sum Java

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.

For 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.

题意及分析:求从根节点到叶子结点上的点相加,和是否有等于sum的。使用递归即可。遍历树,对每遍历一个点就减去用sum减去该点的值,就得到从根节点到当前点剩余的值,若对于叶子节点有sum-node.val==0,那么就存在题目中要求的叶子节点。

代码:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null) return false;
        if(root.left==null&&root.right==null&&sum-root.val==0) return true;
        return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);
    }
}
原文地址:https://www.cnblogs.com/271934Liao/p/7216022.html