Invert Binary Tree

refer to : https://www.algoexpert.io/questions/Invert%20Binary%20Tree

1. problem statement.

swap every left node in the tree for its corresponding right node. 

2. 递归法（深度优先遍历）

O(n) time： we are traversing every single node

O(d) space: 需要存放 O(d) 个函数调用(d是树的深度)

import java.util.*;

class Program {
  //O(n) time | O(d) space
  public static void invertBinaryTree(BinaryTree tree) {
        if(tree ==null){
            return;
        }
        swapHelper(tree);
        invertBinaryTree(tree.left);
        invertBinaryTree(tree.right);
        
  }
    
    public static void swapHelper(BinaryTree tree){
        BinaryTree left = tree.left;
        tree.left = tree.right;
        tree.right = left;
    }

  static class BinaryTree {
    public int value;
    public BinaryTree left;
    public BinaryTree right;

    public BinaryTree(int value) {
      this.value = value;
    }
  }
}

def invertBinaryTree(tree):
    if tree is None:
        return
    swapHelper(tree)
    invertBinaryTree(tree.left)
    invertBinaryTree(tree.right)
    

def swapHelper(tree):
    tree.left, tree.right = tree.right, tree.left


# This is the class of the input binary tree.
class BinaryTree:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

 

3.迭代法（广度优先遍历, 队列）

O(n) time 

O(n) space , 假设到了最后一层，我们把最后一层的所有孩子结点都加进队列中，for balanced tree, n/2 leaf nodes

1. initialize a queue
2. traverse level by level, root node-->next level
3. traverse the tree by using BFS manually
4. pop one node one time, get curr node
5. swap the curr.left and curr.right
6. add curr.left and curr.right into the queue
7. repeat 4-6 until queue is empty.

import java.util.*;

class Program {
    //O(n) time | O(n) space
  public static void invertBinaryTree(BinaryTree tree) {
        ArrayDeque<BinaryTree> queue = new ArrayDeque<BinaryTree>();
        //addLast() method to insert at end 
        queue.addLast(tree);
        while(queue.size() > 0){
            //removes the first element of the Deque and returns the same
            BinaryTree curr = queue.pollFirst();
            swapHelper(curr);
            if(curr.left != null){
                queue.addLast(curr.left);
            }
            if(curr.right != null){
                queue.addLast(curr.right);
            }
        }
  }
    
    public static void swapHelper(BinaryTree tree){
        BinaryTree left = tree.left;
        tree.left = tree.right;
        tree.right = left;
    }

  static class BinaryTree {
    public int value;
    public BinaryTree left;
    public BinaryTree right;

    public BinaryTree(int value) {
      this.value = value;
    }
  }
}

def invertBinaryTree(tree):
    queue = [tree]
    while len(queue):
        curr = queue.pop(0)
        if curr is None:
            continue
        swapHelper(curr)
        queue.append(curr.left)
        queue.append(curr.right)

def swapHelper(tree):
    tree.left, tree.right = tree.right, tree.left
# This is the class of the input binary tree.
class BinaryTree:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None