[LeetCode] 129. Sum Root to Leaf Numbers 解题思路

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.

问题：给定一个二叉树，树的每个节点都只包含一个数字 0-9，将根节点到叶子节点路径上的元素值组合成一个整数，求所有整数和。

这是一道深度遍历的应用。深度遍历二叉树一般是递归遍历，或者借助栈遍历。每到达一个叶子节点，都需要重新遍历一次根节点到该叶子节点路径的值，借组栈遍历可以实现满足这个需要。

    #define null_symble  -2147483648

    int sumNumbers(TreeNode* root) {
        
        if(root == NULL){
            return 0;
        }
        
        list<TreeNode*> stackt;
        stackt.push_back(root);
        
        int sum = 0;

        map<TreeNode*, int> node_val;

        while(stackt.size() > 0){
            TreeNode* node = stackt.back();
            
            if(node->val != null_symble){
                node_val[node] = node->val;
                node->val = null_symble;
            }
            
            if (node->left != NULL && node->left->val != null_symble){
                stackt.push_back(node->left);
                continue;
            }
            
            if(node->right != NULL && node->right->val != null_symble){
                stackt.push_back(node->right);
                continue;
            }
            
            if(node->left == NULL && node->right == NULL){
                string str = "";
                list<TreeNode*>::iterator l_iter;
                for(l_iter = stackt.begin(); l_iter != stackt.end(); l_iter++){
                    str += to_string(node_val[*l_iter]);
                }
                sum += stoi(str);
            }
            stackt.pop_back();
        }
        
        return sum;
    }