leetcode 653. Two Sum IV - Input is a BST

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

Example 1:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 9

Output: True

Example 2:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 28

Output: False

解法1:使用dfs还原tree为一个sorted array，然后使用two sum算法求解。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def findTarget(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: bool
        """        
        # use DFS, transform tree to ordered array.       
        def dfs(node, stack):
            if not node: return
            dfs(node.left, stack)
            stack.append(node.val)
            dfs(node.right, stack)
            
        stack = []
        dfs(root, stack)
        
        i, j = 0, len(stack)-1
        while i < j:
            if stack[i] + stack[j] == k:
                return True
            elif stack[i] + stack[j] > k:
                j -= 1
            else:
                i += 1
        return False                

解法2:

上述算法使用迭代，

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def findTarget(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: bool
        """        
        # use DFS, transform tree to ordered array.       
        def dfs(node, arr):
            if not node: return
            stack = []
            while node:
                stack.append(node)
                node = node.left
            while stack:
                node = stack.pop()
                arr.append(node.val)
                node = node.right
                while node:
                    stack.append(node)
                    node = node.left            
            
        stack = []
        dfs(root, stack)
        
        i, j = 0, len(stack)-1
        while i < j:
            if stack[i] + stack[j] == k:
                return True
            elif stack[i] + stack[j] > k:
                j -= 1
            else:
                i += 1
        return False                

解法3:

迭代tree，使用hash：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def findTarget(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: bool
        """        
        # use DFS, hash record accessed node.
        arr = set()
        def dfs(node, arr):
            if not node: return False          
            if k-node.val in arr:
                return True
            arr.add(node.val)
            return dfs(node.left, arr) or dfs(node.right, arr) 
        return dfs(root, arr)
                       

 

Method 4.
The idea is to use binary search method. For each node, we check if k - node.val exists in this BST.

Time Complexity: O(nlogn), Space Complexity: O(h). h is the height of the tree, which is logn at best case, and n at worst case.

Java version:

    public boolean findTarget(TreeNode root, int k) {
        return dfs(root, root,  k);
    }
    
    public boolean dfs(TreeNode root,  TreeNode cur, int k){
        if(cur == null)return false;
        return search(root, cur, k - cur.val) || dfs(root, cur.left, k) || dfs(root, cur.right, k);
    }
    
    public boolean search(TreeNode root, TreeNode cur, int value){
        if(root == null)return false;
        return (root.val == value) && (root != cur) 
            || (root.val < value) && search(root.right, cur, value) 
                || (root.val > value) && search(root.left, cur, value);
    }

 

解法5：从min到max迭代tree node，从max 到min迭代tree node：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def findTarget(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: bool
        """        
        # iterate tree node from min to max
        # iterate tree node from max to min
        # check two node of sum == k
        
        def add_node(stack, node, is_left=True):        
            while node:
                stack.append(node)
                node = node.left if is_left else node.right
        
        if not root: return False 
        
        stack1, stack2 = [], []
        add_node(stack1, root, True)    
        add_node(stack2, root, False)    
        
        while stack1 and stack2:
            node1 = stack1[-1]
            node2 = stack2[-1]            
            if node1 is node2: return False
            if node1.val + node2.val == k:
                return True
            elif node1.val + node2.val > k:
                node2 = stack2.pop()
                add_node(stack2, node2.left, False)                                   
            else:
                node1 = stack1.pop()
                add_node(stack1, node1.right, True)                  
        return False