【LeetCode】129. 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.

题解：

　　类似于 Path Sum 和  Path Sum II，迭代方法依然是后序遍历的思路，先遍历左右孩子。

Solution 1

 class Solution {
 public:
     int sumNumbers(TreeNode* root) {
         if (!root)
             return ;
         return dfs(root, );
     }
     int dfs(TreeNode* root, int sum) {
         if (!root)
             return ;
         sum = sum *  + root->val;
         if (!root->left && !root->right)
             return sum;
         return dfs(root->left, sum) + dfs(root->right, sum);
     }
 };

Solution 2

 class Solution {
 public:
     int sumNumbers(TreeNode* root) {
         if (!root)
             return ;
         queue<TreeNode*> q1;
         queue<int> q2;
         q1.push(root);
         q2.push(root->val);
 
         int sum = , cursum = ;
         TreeNode* cur = root;
 
         while (!q1.empty()) {
             cur = q1.front();
             cursum = q2.front();
             q1.pop();
             q2.pop();
 
             if (!cur->left && !cur->right) {
                 sum += cursum;
             }
             if (cur->left) {
                 q1.push(cur->left);
                 q2.push(cursum *  + cur->left->val);
             }
             if (cur->right) {
                 q1.push(cur->right);
                 q2.push(cursum *  + cur->right->val);
             }
         }
         return sum;
     }
 };

Solution 3

 class Solution {
 public:
     int sumNumbers(TreeNode* root) {
         if (!root)
             return ;
         stack<TreeNode*> s1;
         stack<int> s2;
         int sum = , cursum = ;
         TreeNode* cur = root, *pre = nullptr;
         while (cur || !s1.empty()) {
             while (cur) {
                 s1.push(cur);
                 cursum = cursum *  + cur->val;
                 s2.push(cursum);
                 cur = cur->left;
             }
             cur = s1.top();
             if (!cur->left && !cur->right) {
                 sum += cursum;
             }
             if (cur->right && cur->right != pre) {
                 cur = cur->right;
             } else {
                 pre = cur;
                 s1.pop();
                 cursum = s2.top();
                 s2.pop();
                 cursum -= cur->val;
                 cursum /= ;
                 cur = nullptr;
             }
         }
         return sum;
     }
 };