add prob #1022; iterative solution

Author: randomwalk10

1022_Sum_of_Root_To_Leaf_Binary_Numbers.cpp

@@ -0,0 +1,60 @@
+/*
+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.
+
+Note:
+
+The number of nodes in the tree is between 1 and 1000.
+node.val is 0 or 1.
+The answer will not exceed 2^31 - 1.
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/sum-of-root-to-leaf-binary-numbers
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+*/
+
+#include <stack>
+using namespace std;
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+};
+
+class Solution {
+public:
+    int sumRootToLeaf(TreeNode* root) {
+		if(NULL==root) return 0;
+		int res = 0;
+		stack<TreeNode*> s_parent;
+		stack<int> s_sum;
+		
+		s_parent.push(root);
+		s_sum.push(0);
+		while(!s_parent.empty()){
+			TreeNode* tmp_node = s_parent.top(); s_parent.pop();
+			int tmp_sum = s_sum.top(); s_sum.pop();
+			tmp_sum = (tmp_sum<<1) + tmp_node->val;
+
+			if((tmp_node->left==NULL)&&(tmp_node->right==NULL)){
+				res += tmp_sum;
+				continue;
+			}
+			if(tmp_node->right){
+				s_parent.push(tmp_node->right);
+				s_sum.push(tmp_sum);
+			}
+			if(tmp_node->left){
+				s_parent.push(tmp_node->left);
+				s_sum.push(tmp_sum);
+			}
+		}
+
+		return res;
+    }
+};