Path Sum

来源：互联网发布：安卓tv软件编辑：程序博客网时间：2024/05/24 06:46

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.

确认是否有等于指定值的路径和只要计算根节点到每个叶子的路径值之和即可采用递归记录到达每个节点时的和判断该节点是否为叶子是则进行判断不是则继续递归代码如下：

public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        return pathSum(root,sum,0);    }    public boolean pathSum(TreeNode root,int sum,int cur){        if(root==null)return false;        if(root.left==null&&root.right==null)            return sum==cur+root.val? true:false;        if(root.left==null)return pathSum(root.right,sum,cur+root.val);        else{            if(root.right==null)return pathSum(root.left,sum,cur+root.val);            else return pathSum(root.left,sum,cur+root.val)||pathSum(root.right,sum,cur+root.val);        }    }}

0 0