[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)

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)

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:

# 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:

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++:

// 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++:

// 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;
    }
};