Leetcode 156:binary Tree Upside down

Binary Tree Upside DownTotal Accepted: 813 Total Submissions: 2515

Given a binary tree where all the right nodes is either leaf nodes with a sibling (a left node that shares the same paren T node) or empty, flip it upside down and turn it to a tree where the original right nodes turned to 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  

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Analysis

Understand the definition of this tree, the answer is self-evident.

Note

None

[CODE]

/** * Definition for Binary tree * public class TreeNode {*     int val, *     TreeNode left, *     TreeNode right; *
   
    treenode (int x) {val = x;} *} */public class Solution {public    TreeNode upsidedownbinarytree (TreeNode root) {        stack<treenode> stack = new stack<treenode> ();                if (root==null) return null;        while (root.left! = null) {Stack.push (root); root=root.left;}        Stack.push (root);        while (!stack.empty ()) {            TreeNode node = Stack.pop ();            if (!stack.empty ()) {                node.right = Stack.peek ();                Node.left = Stack.peek (). right;            } else {                node.left = null;                Node.right = null;            }        }        return root;    }}
   

Leetcode 156:binary Tree Upside down