Leetcode dfs Sum Root to Leaf Numbers

Sum Root to Leaf Numbers

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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.

题意：给定一棵二叉树，每个节目包含数字0-9，根到叶子节点组成一个数字，

求所有这些数字的总和

思路：dfs

int sum

void dfs(TreeNode *root, int num)

num表示从根节点到当前节点前面的节点组成数字

这个函数求出，在num的基础上，从root 为根到它的所有叶子节点组成的数字的总和

复杂度：时间 O(n) ；空间 O(log n)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

    int sumNumbers(TreeNode *root){
    	return dfs(root, 0);
    }
private:
    int dfs(TreeNode *root, int num){
    	if(!root) {return 0;} //空
    	if(!root->left && !root->right) {return num * 10 + root->val;} //叶子节点
    	return dfs(root->left, num * 10 + root->val) + dfs(root->right, num * 10 + root->val);
    }

};