Two Sum BSTs

Given two binary search trees, return True if and only if there is a node in the first tree and a node in the second tree whose values sum up to a given integer target.

 

Example 1:

 

Input: root1 = [2,1,4], root2 = [1,0,3], target = 5
Output: true
Explanation: 2 and 3 sum up to 5.

思路：就是个hashset存target - node.val, 两个inorder traverse；

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        if(root1 == null || root2 == null) {
            return false;
        }
        HashSet<Integer> set = new HashSet<Integer>();
        traverse(root1, set, target, true);
        return traverse(root2, set, target, false);
    }
    
    private boolean traverse(TreeNode root,  HashSet<Integer> set, 
                             int target, boolean traverse) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode node = root;
        while(node != null) {
            stack.push(node);
            node = node.left;
        }
        
        while(!stack.isEmpty()) {
            node = stack.pop();
            if(traverse) {
                set.add(target - node.val);
            } else {
                if(set.contains(node.val)) {
                    return true;
                }
            }
            if(node.right != null) {
                node = node.right;
                while(node != null) {
                    stack.push(node);
                    node = node.left;
                }
            }
        }
        return false;
    }
}

思路2：用两个stack，stack1走到最左边，stack2走到最右边，然后sum < target, 走左边，sum>target走右边；O(M + N);

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        if(root1 == null || root2 == null) {
            return false;
        }
        Stack<TreeNode> stack1 = new Stack<TreeNode>();
        Stack<TreeNode> stack2 = new Stack<TreeNode>();
        
        TreeNode t1, t2;
        
        while(true) {
            while(root1 != null) {
                stack1.push(root1);
                root1 = root1.left;
            }
            
            while(root2 != null) {
                stack2.push(root2);
                root2 = root2.right;
            }
            if(stack1.isEmpty() || stack2.isEmpty()) {
                break;
            }
            
            t1 = stack1.peek();
            t2 = stack2.peek();
            
            int sum = t1.val + t2.val;
            if(sum == target) {
                return true;
            } else if(sum < target) {
                stack1.pop();
                root1 = t1.right;
            } else {
                // sum > target;
                stack2.pop();
                root2 = t2.left;
            }
        }
        return false;
    }
}