[LeetCode] 156. Binary Tree Upside Down 二叉树的上下颠倒

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:

Given a binary tree {1,2,3,4,5},

1

/ \

2 3

/ \

4 5

return the root of the binary tree [4,5,2,#,#,3,1].

4

/ \

5 2

/ \

3 1

 

给一个二叉树，右节点要么为空要么一定会有对应的左节点，把二叉树上下颠倒一下，原二叉树的最左子节点变成了根节点，其对应的右节点变成了其左子节点，其父节点变成了其右子节点。

解法1：递归

解法2：迭代

Java: Time: O(N), Space: O(N)

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public class Solution {     public TreeNode upsideDownBinaryTree(TreeNode root) {         if(root == null || root.left == null)return root;         TreeNode newRoot = upsideDownBinaryTree(root.left);         //root.left is newRoot everytime         root.left.left = root.right;         root.left.right = root;         root.left = null;         root.right = null;         return newRoot;     } }

Java: Time: O(N), Space: O(1)

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public class Solution {     public TreeNode upsideDownBinaryTree(TreeNode root) {         TreeNode cur = root;         TreeNode pre = null;         TreeNode tmp = null;         TreeNode next = null;         while(cur != null){             next = cur.left;             //need tmp to keep the previous right child             cur.left = tmp;             tmp = cur.right;                           cur.right = pre;             pre = cur;             cur = next;         }         return pre;     } }

Python:

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# Time:  O(n) # Space: O(n) class Solution2(object):     # @param root, a tree node     # @return root of the upside down tree     def upsideDownBinaryTree(self, root):         return self.upsideDownBinaryTreeRecu(root, None)       def upsideDownBinaryTreeRecu(self, p, parent):         if p is None:             return parent           root = self.upsideDownBinaryTreeRecu(p.left, p)         if parent:             p.left = parent.right         else:             p.left = None         p.right = parent           return root

Python:

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class Solution(object):     # @param root, a tree node     # @return root of the upside down tree     def upsideDownBinaryTree(self, root):         p, parent, parent_right = root, None, None           while p:             left = p.left             p.left = parent_right             parent_right = p.right             p.right = parent             parent = p             p = left           return parent

C++:

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// Recursion class Solution { public:     TreeNode *upsideDownBinaryTree(TreeNode *root) {         if (!root || !root->left) return root;         TreeNode *l = root->left, *r = root->right;         TreeNode *res = upsideDownBinaryTree(l);         l->left = r;         l->right = root;         root->left = NULL;         root->right = NULL;         return res;     } };

C++:

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// Iterative class Solution { public:     TreeNode *upsideDownBinaryTree(TreeNode *root) {         TreeNode *cur = root, *pre = NULL, *next = NULL, *tmp = NULL;         while (cur) {             next = cur->left;             cur->left = tmp;             tmp = cur->right;             cur->right = pre;             pre = cur;             cur = next;         }         return pre;     } };