129.Sum Root to Leaf Numbers

Tags: Medium Depth-first Search Tree

Links: https://leetcode.com/problems/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.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

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/**
 * 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 sumNumbers(TreeNode* root) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return 0;

        if (!root -> left && !root -> right) return root -> val;
        int l = 0, r = 0;
        if (root -> left) {
            root -> left -> val += 10 * root -> val;
            l = sumNumbers(root -> left);
        }
        if (root -> right) {
            root -> right -> val += 10 * root -> val;
            r = sumNumbers(root -> right);
        }
        return l + r;
    }
};

迭代解法，相当于先序遍历。

/**
 * 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 sumNumbers(TreeNode* root) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return 0;

        int res = 0;
        stack<TreeNode *> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode *tmp = s.top(); s.pop();
            if (!tmp -> left && !tmp -> right)
                res += tmp -> val;
            if (tmp -> right) {
                tmp -> right -> val += tmp -> val * 10;
                s.push(tmp -> right);
            }
            if (tmp -> left) {
                tmp -> left -> val += tmp -> val * 10;
                s.push(tmp -> left);
            }
        }

        return res;
    }
};