树——sum-root-to-leaf-numbers
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题目:
二叉树每个结点包含0-9的数字,例如一条root到leaf的路径为h1->2->3,则该路径和为123,求该二叉树总的路径和。
,
For example,
1 / \ 2 3
The root-to-leaf path1->2represents the number12.
The root-to-leaf path1->3represents the number13.
Return the sum = 12 + 13 =25.
思路:穷尽遍历二叉树,每次遍历到叶节点就将该路径的和加入到totalsum中,最后返回totalsum.
代码如下:
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */public class Solution { private int totalsum=0; public int sumNumbers(TreeNode root) { if(root == null) return 0; pathsum(root,0); return totalsum; } public void pathsum(TreeNode root,int presum)//presum为该节点之前路径的和; { if(root.left == null&&root.right == null) totalsum+=presum*10+root.val;//每次遍历到根节点,将该路径和加在totalsum上; if(root.left!=null) pathsum(root.left,presum*10+root.val); if(root.right!=null) pathsum(root.right,presum*10+root.val); }}
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