LeetCode 23 Sum Root to left Numbers_leetcode root.left-优快云博客

本文介绍了一种解决特定二叉树问题的方法：计算所有从根节点到叶节点的路径所表示数字的总和。通过递归算法实现，详细解释了如何在遍历过程中累积路径数值，并最终求得所有路径数值之和。

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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.

分析：二叉树问题，一般来讲要首先要想到递归。

有一个num，沿着路径一直num = num*10+node.val

有一个sum，碰见叶子节点，则sum = sum + num

/**
 * 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) {
        if(root==null) return 0;
        return sum(root, 0, 0);
    }
    public int sum(TreeNode node, int num, int sum){
        if(node==null) return sum;
        //每碰见一个节点，就更新num
        num = num*10+node.val;
        if(node.left==null && node.right==null){
            //遇到叶节点，则更新sum
            sum = sum + num;
            return sum;
        }
        //递归地，将两条路径加起来
        sum = sum(node.left, num, sum) + sum(node.right, num, sum);
        return sum;
    }
}