[LeetCode] 114. Flatten Binary Tree to Linked List 将二叉树展平为链表

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / \
  2   5
 / \   \
3   4   6

The flattened tree should look like:

1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

 

给一个二叉树，把它展平为链表 in-place

根据展平后的链表的顺序可以看出是先序遍历的结果，所以用inorder traversal。

解法：递归

解法：迭代

Java:

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/**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode(int x) { val = x; }  * }  */ class Solution {     private TreeNode prev = null;       public void flatten(TreeNode root) {         if (root == null)             return;         flatten(root.right);         flatten(root.left);         root.right = prev;         root.left = null;         prev = root;     } }

Java:

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public void flatten(TreeNode root) {         if (root == null) return;         Stack<TreeNode> stk = new Stack<TreeNode>();         stk.push(root);         while (!stk.isEmpty()){             TreeNode curr = stk.pop();             if (curr.right!=null)                   stk.push(curr.right);             if (curr.left!=null)                   stk.push(curr.left);             if (!stk.isEmpty())                  curr.right = stk.peek();             curr.left = null;  // dont forget this!!         }     }

Python:

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# Definition for a  binary tree node class TreeNode:     def __init__(self, x):         self.val = x         self.left = None         self.right = None   class Solution:     # @param root, a tree node     # @return nothing, do it in place     def flatten(self, root):         return self.flattenRecu(root, None)       def flattenRecu(self, root, list_head):         if root != None:             list_head = self.flattenRecu(root.right, list_head)             list_head = self.flattenRecu(root.left, list_head)             root.right = list_head             root.left = None             return root         else:             return list_head

Python:

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class Solution:     list_head = None     # @param root, a tree node     # @return nothing, do it in place     def flatten(self, root):         if root != None:             self.flatten(root.right)             self.flatten(root.left)             root.right = self.list_head             root.left = None             self.list_head = root             return root

C++:

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// Recursion class Solution { public:     void flatten(TreeNode *root) {         if (!root) return;         if (root->left) flatten(root->left);         if (root->right) flatten(root->right);         TreeNode *tmp = root->right;         root->right = root->left;         root->left = NULL;         while (root->right) root = root->right;         root->right = tmp;     } };

C++:

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class Solution { public:     void flatten(TreeNode* root) {         if (!root) return;         stack<TreeNode*> s;         s.push(root);         while (!s.empty()) {             TreeNode *t = s.top(); s.pop();             if (t->left) {                 TreeNode *r = t->left;                 while (r->right) r = r->right;                 r->right = t->right;                 t->right = t->left;                 t->left = NULL;             }             if (t->right) s.push(t->right);         }     } };

　　

　　

　　

　　

 

 

 

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