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https://leetcode.com/problems/sum-root-to-leaf-numbers/
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
For example,
1 / 2 3
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Return the sum = 12 + 13 = 25.
解题思路：
比较典型的DFS+回溯，结合二叉树的遍历。这里递归要传入当前路径的和，以及总路径的和。但是由于java里面只能pass by value，而不能pass by reference，一般可以通过return一个值来解决。这里通过传入一个int[]的对象，就可以伪装成pass by reference了。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public int sumNumbers(TreeNode root) {
int[] sum = new int[2]; dfs(root, sum); return sum[0]; } public void dfs(TreeNode root, int[] sum){ if(root == null){ return; } if(root.left == null && root.right == null){ sum[1] = sum[1] * 10 + root.val; sum[0] += sum[1]; sum[1] = sum[1] / 10; //重要，这里的回溯不能忘记 return; } sum[1] = sum[1] * 10 + root.val; dfs(root.left, sum); dfs(root.right, sum); sum[1] = sum[1] / 10; } }
本题还有另一种递归思路。一棵树的path sum，就是它左右子树的path sum。如此递归。
这里需要传入从root到当前节点的和，用以往后继续计算。特别要注意的，如果当前节点为null，应该返回0，而不是返回pathSum。因为看最后的递归返回，这代表当前子树的和为0。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public int sumNumbers(TreeNode root) { return dfs(root, 0); } public int dfs(TreeNode root, int pathSum){ if(root == null){ return 0; //不是return pathSum; } pathSum = pathSum * 10 + root.val; if(root.left == null && root.right == null){ return pathSum; } return dfs(root.left, pathSum) + dfs(root.right, pathSum); } }
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原文地址：http://www.cnblogs.com/NickyYe/p/4350918.html
