Path Sum

标签：sum   path   tree   递归   

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.

              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);
        }
    }
}

Path Sum

标签：sum   path   tree   递归   

原文地址：http://blog.csdn.net/u012734829/article/details/42268605