[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:

/**
 * 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:

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:

# 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:

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++:

// 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++:

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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