recoverTreeFromPreorderTraversal.cpp

algorithms/cpp/recoverBinarySearchTree/recoverTreeFromPreorderTraversal.cpp

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+// source : https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/
+// Author Ahmed Morsy
+// Date   : 2019-5-19
+
+/********************************************************************************** 
+*We run a preorder depth first search on the root of a binary tree.
+*At each node in this traversal, we output D dashes (where D is the depth of this node), 
+*then we output the value of this node.  
+*(If the depth of a node is D, the depth of its immediate child is D+1.The depth of the root node is 0.)
+*If a node has only one child, that child is guaranteed to be the left child.
+*Given the output S of this traversal, recover the tree and return its root.
+
+* Input: "1-2--3--4-5--6--7"
+* Output: [1,2,5,3,4,6,7]
+
+* Proposed Solution
+  -----------------
+ 1. for each node in the input save its depth and value
+ 2. start a stack with the root node pushed onto it
+
+    IF the following node has the root's depth + 1 and the root has less than 2 children then this node
+    is a child for this root. Add it to the stack.
+
+    ELSE then pop the current node from the stack.
+
+**********************************************************************************/
+
+/*  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:
+
+    TreeNode* recoverFromPreorder(string S) {
+        vector<int>values,depth;
+        int cur_val = 0 , cur_depth = 0;
+        bool dash = false;
+        for(char s : S){
+            if(s == '-'){
+                if(!dash){
+                    values.push_back(cur_val);
+                    depth.push_back(cur_depth);
+                    cur_depth = 0;
+                    cur_val = 0;
+                }
+                dash = true;
+                cur_depth++;
+            }
+            else{
+                dash = false;
+                cur_val *= 10;
+                cur_val += s-'0';
+            }
+        }
+        values.push_back(cur_val);
+        depth.push_back(cur_depth);
+
+        unordered_map<TreeNode*,int>depths;
+
+
+        int ptr = 1;
+        TreeNode *root = new TreeNode(values[0]);
+        depths[root] = 0;
+        stack<TreeNode*>st;
+        st.push(root);
+
+        while(ptr < (int)values.size()){
+            TreeNode *cur = st.top();
+            if(depth[ptr] == depths[cur]+1 && (cur->left == NULL || cur->right == NULL)){
+                TreeNode *t = new TreeNode(values[ptr++]);
+                depths[t] = depths[cur]+1;
+                if(cur->left == NULL){
+                    cur->left = t;
+                }
+                else{
+                    cur->right = t;
+                }
+                st.push(t);
+            }
+            else{
+                st.pop();
+            }
+        }
+        return root;
+
+    }
+};
+