[Leetcode 1022] Sum of Root To Leaf Binary Numbers

原题说明

Given a binary tree, each node has value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit. For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.

Return the sum of these numbers.

Example 1:

Input: [1,0,1,0,1,0,1]
Output: 22
Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22

Note:

1. The number of nodes in the tree is between 1 and 1000.
2. node.val is 0 or 1.
3. The answer will not exceed 2^31 - 1.

解题思路

这题比较简单，通过dfs先序遍历二叉树，分别计算每一条从root到leaf的二进制数，累加即可。

示例代码 (cpp)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumRootToLeaf(TreeNode* root) {
        int ans = 0;
        helperFunc(root, ans);
        return ans;
    }
    void helperFunc(TreeNode* Node, int& ans, int tmpSum = 0) {
        tmpSum = tmpSum * 2 + Node->val;
        if (Node->right) {
            helperFunc(Node->right, ans, tmpSum);
        }
        if (Node->left) {
            helperFunc(Node->left, ans, tmpSum);
        }
        if (!Node->right && !Node->left) {
            ans += tmpSum;
        }
    }
};

复杂度分析

时间复杂度: O(n)
空间复杂度: O(1)

归纳总结

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