[LeetCode][114. Flatten Binary Tree to Linked List] 3 Approaches: PreOrder, Iteration and Recursion

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By Long Luo

This article is the solution Image Explanation to Understand the Recursion Solution of Problem 114. Flatten Binary Tree to Linked List .

The Binary Tree Traversal Algorithms can be find here Tree Traversal Algorithms: PreOrder, InOrder and PostOrder .

We can use DFS to traversal the binary tree.

PreOrder + List

It’s easy to find that the flatten tree is the PreOrder of the tree. We can preorder the tree and store the TreeNodes in a List.

Traversal the list, make the current node as the preNode’s right node.

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public void flatten(TreeNode root) {
    if (root == null) {
        return;
    }

    List<TreeNode> nodeList = new ArrayList<>();
    preOrder(root, nodeList);
    int n = nodeList.size();
    for (int i = 1; i < n; i++) {
        TreeNode preNode = nodeList.get(i - 1);
        TreeNode currNode = nodeList.get(i);
        preNode.left = null;
        preNode.right = currNode;
    }
}

public void preOrder(TreeNode root, List<TreeNode> nodeList) {
    if (root == null) {
        return;
    }

    nodeList.add(root);
    preOrder(root.left, nodeList);
    preOrder(root.right, nodeList);
}

We can also traversal the tree iteratively with a Stack:

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public void flatten(TreeNode root) {
    if (root == null) {
        return;
    }

    List<TreeNode> nodeList = new ArrayList<>();
    Stack<TreeNode> stk = new Stack<>();
    while (root != null || !stk.empty()) {
        while (root != null) {
            nodeList.add(root);
            stk.push(root);
            root = root.left;
        }

        root = stk.pop();
        root = root.right;
    }

    int len = nodeList.size();
    for (int i = 1; i < len; i++) {
        TreeNode preNode = nodeList.get(i - 1);
        TreeNode currNode = nodeList.get(i);
        preNode.left = null;
        preNode.right = currNode;
    }
}

Analysis

  • Time Complexity: \(O(n)\).
  • Space Complexity: \(O(n)\).

Iteration

The Process is as follows:

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    1
   / \
  2   5
 / \   \
3   4   6

// 1. Insert the left subtree of 1 to the right subtree
    1
     \
      2         5
     / \         \
    3   4         6        

// 2. Let the right subtree to the right node of the farest right node of the left subtree.  
    1
     \
      2          
     / \          
    3   4  
         \
          5
           \
            6
            
 //3. Insert the left subtree of 2 to the right subtree
    1
     \
      2          
       \          
        3       4  
                 \
                  5
                   \
                    6   
        
 // 4. Let the original right subtree to the right node of the farest right node of the left subtree.
    1
     \
      2          
       \          
        3      
         \
          4  
           \
            5
             \
              6         
  
  ......

Let’s coding it.

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public void flatten(TreeNode root) {
    while (root != null) {
        if (root.left == null) {
            root = root.right;
        } else {
            TreeNode pre = root.left;
            while (pre.right != null) {
                pre = pre.right;
            }

            pre.right = root.right;
            root.right = root.left;
            root.left = null;
            root = root.right;
        }
    }
}

Analysis

  • Time Complexity: \(O(n)\)
  • Space Complexity: \(O(n)\)

Recursion

To Flatten Binary Tree to Linked List, there are 3 steps as the picture shows.

DFS
  1. Flatten the left subtree of the root node into a linked list;
  2. Flatten the right subtree of the root node into a linked list;
  3. Let the right subtree of the step 2 be the right child of the farest right node of the left subtree of step 1.

Obiously, that’s a recursion process.

Let’s coding it.

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  public static void flatten_rec(TreeNode root) {
      if (root == null) {
          return;
      }

// left subtree
      flatten_rec(root.left);
// right subtree
      flatten_rec(root.right);

      TreeNode temp = root.right;
      root.right = root.left;
      root.left = null;

      // find the farest right node.
      while (root.right != null) {
          root = root.right;
      }

      root.right = temp;
  }

Analysis

  • Time Complexity: \(O(n)\).
  • Space Complexity: \(O(n)\).

References:

  1. 详细通俗的思路分析，多解法

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