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1028. Recover a Tree From Preorder Traversal

Description

We run a preorder depth-first search (DFS) 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 traversal of this traversal, recover the tree and return its root.

 

Example 1:

Input: traversal = "1-2--3--4-5--6--7"
Output: [1,2,5,3,4,6,7]

Example 2:

Input: traversal = "1-2--3---4-5--6---7"
Output: [1,2,5,3,null,6,null,4,null,7]

Example 3:

Input: traversal = "1-401--349---90--88"
Output: [1,401,null,349,88,90]

 

Constraints:

• The number of nodes in the original tree is in the range [1, 1000].
• 1 <= Node.val <= 109

Solutions

• C++
• Java
• TypeScript
• Javascript

• /**
   * 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) {
          stack<TreeNode*> st;
          int depth = 0;
          int num = 0;
          for (int i = 0; i < S.length(); ++i) {
              if (S[i] == '-') {
                  depth++;
              } else {
                  num = 10 * num + S[i] - '0';
              }
              if (i + 1 >= S.length() || (isdigit(S[i]) && S[i + 1] == '-')) {
                  TreeNode* newNode = new TreeNode(num);
                  while (st.size() > depth) {
                      st.pop();
                  }
                  if (!st.empty()) {
                      if (st.top()->left == nullptr) {
                          st.top()->left = newNode;
                      } else {
                          st.top()->right = newNode;
                      }
                  }
                  st.push(newNode);
                  depth = 0;
                  num = 0;
              }
          }
          TreeNode* res;
          while (!st.empty()) {
              res = st.top();
              st.pop();
          }
          return res;
      }
  };
  

• /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode() {}
   *     TreeNode(int val) { this.val = val; }
   *     TreeNode(int val, TreeNode left, TreeNode right) {
   *         this.val = val;
   *         this.left = left;
   *         this.right = right;
   *     }
   * }
   */
  class Solution {
      public TreeNode recoverFromPreorder(String traversal) {
          Stack<TreeNode> stack = new Stack<>();
          int i = 0;
  
          while (i < traversal.length()) {
              int depth = 0;
              while (i < traversal.length() && traversal.charAt(i) == '-') {
                  depth++;
                  i++;
              }
  
              int num = 0;
              while (i < traversal.length() && Character.isDigit(traversal.charAt(i))) {
                  num = num * 10 + (traversal.charAt(i) - '0');
                  i++;
              }
  
              // Create the new node
              TreeNode newNode = new TreeNode(num);
  
              while (stack.size() > depth) {
                  stack.pop();
              }
              if (!stack.isEmpty()) {
                  if (stack.peek().left == null) {
                      stack.peek().left = newNode;
                  } else {
                      stack.peek().right = newNode;
                  }
              }
  
              stack.push(newNode);
          }
          return stack.isEmpty() ? null : stack.get(0);
      }
  }
  
  

• function recoverFromPreorder(traversal: string): TreeNode | null {
      const stack: TreeNode[] = [];
      let i = 0;
  
      while (i < traversal.length) {
          let depth = 0;
          while (i < traversal.length && traversal[i] === '-') {
              depth++;
              i++;
          }
  
          let num = 0;
          while (i < traversal.length && !Number.isNaN(+traversal[i])) {
              num = num * 10 + +traversal[i];
              i++;
          }
  
          // Create the new node
          const newNode = new TreeNode(num);
  
          while (stack.length > depth) {
              stack.pop();
          }
  
          if (stack.length > 0) {
              const i = stack.length - 1;
              if (stack[i].left === null) {
                  stack[i].left = newNode;
              } else {
                  stack[i].right = newNode;
              }
          }
  
          stack.push(newNode);
      }
  
      return stack.length ? stack[0] : null;
  }
  
  

• function recoverFromPreorder(traversal) {
      const stack = [];
      let i = 0;
  
      while (i < traversal.length) {
          let depth = 0;
          while (i < traversal.length && traversal[i] === '-') {
              depth++;
              i++;
          }
  
          let num = 0;
          while (i < traversal.length && !Number.isNaN(+traversal[i])) {
              num = num * 10 + +traversal[i];
              i++;
          }
  
          // Create the new node
          const newNode = new TreeNode(num);
  
          while (stack.length > depth) {
              stack.pop();
          }
  
          if (stack.length > 0) {
              const i = stack.length - 1;
              if (stack[i].left === null) {
                  stack[i].left = newNode;
              } else {
                  stack[i].right = newNode;
              }
          }
  
          stack.push(newNode);
      }
  
      return stack.length ? stack[0] : null;
  }
  
  

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