BST_traverse(中序遍历，前序遍历，后序遍历)

refer to: https://www.algoexpert.io/questions/BST%20Traversal && leetcode &&力扣

分别使用前序遍历中序遍历后序遍历的方法遍历搜索二叉树

import java.util.*;

class Program {
  public static List<Integer> inOrderTraverse(BST tree, List<Integer> array) {
        if(tree.left != null){
            inOrderTraverse(tree.left, array);
        }
        array.add(tree.value);
        if(tree.right != null){
            inOrderTraverse(tree.right, array);
        }
        return array;
  }

  public static List<Integer> preOrderTraverse(BST tree, List<Integer> array) {
        array.add(tree.value);
        if(tree.left != null){
            preOrderTraverse(tree.left, array);
        }
        if(tree.right != null){
            preOrderTraverse(tree.right, array);
        }
        return array;
  }

  public static List<Integer> postOrderTraverse(BST tree, List<Integer> array) {
        if(tree.left != null){
            postOrderTraverse(tree.left, array);
        }
        if(tree.right != null){
            postOrderTraverse(tree.right, array);
        }
        array.add(tree.value);
        return array;
    }

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

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

def inOrderTraverse(tree, array):
    # Write your code here.
    if tree is not None:
        inOrderTraverse(tree.left, array)
        array.append(tree.value)
        inOrderTraverse(tree.right, array)
    return array


def preOrderTraverse(tree, array):
    # Write your code here.
    if tree is not None:
        array.append(tree.value)
        preOrderTraverse(tree.left, array)
        preOrderTraverse(tree.right, array)
    return array


def postOrderTraverse(tree, array):
    # Write your code here.
    if tree is not None:
        postOrderTraverse(tree.left, array)
        postOrderTraverse(tree.right, array)
        array.append(tree.value)
    return array

refer to : leedcode && 力扣

import java.util.*;

class Program {
  public static List<Integer> inOrderTraverse(BST tree, List<Integer> array) {
   
        Stack<BST> stack = new Stack<>();
        if(tree == null)   return array;
        BST cur = tree;
        while(cur != null || !stack.isEmpty()){
            while(cur != null){
                stack.push(cur);
                cur = cur.left;
            }
            
            cur = stack.pop();
            array.add(cur.value);
            cur = cur.right;
        }
        return array;
  }

  public static List<Integer> preOrderTraverse(BST tree, List<Integer> array) {
        Stack<BST> stack = new Stack<>();
        if(tree == null)   return array;
        stack.push(tree);
        while(!stack.isEmpty()){
            BST node = stack.pop();
            array.add(node.value);
            if(node.right != null){
                stack.push(node.right);
            }
            if(node.left != null){
                stack.push(node.left);
            }
        }
        return array;
  }
     --why push node.right first, node,left later?  --stack property: first in last out, so last left in, first left out.

  public static List<Integer> postOrderTraverse(BST tree, List<Integer> array) {
    Stack<BST> s = new Stack<>();
    Set<BST> seen = new HashSet<>();
    while (tree != null || !s.isEmpty()) {
       if (tree == null && seen.contains(s.peek())) {
           array.add(s.pop().value);
       } 
       else if (tree == null) {     //step2: 达到了left most,seen.add(left_most node)
           seen.add(s.peek());
           tree = s.peek().right;  //tree=left_most node的右孩子
       } 
       else { //step1 ： tree == null，程序的开始进入这一步,左孩子全部加进栈，直到达到left most node
           s.push(tree);
           tree = tree.left;
       }
    }
    return array;
  }

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

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

time cpmplexity: O(N).   ||     space complexity: O(N)