2019-10-17 leetcode刷题总结 BST

1. 1038. Binary Search Tree to Greater Sum Tree
         根据题目给出的例子，相当于要做一个reverse sort，由于得到sorted array的是inorder traversal，就把inorder sort反过来实现。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode bstToGst(TreeNode root) {
        Stack stack = new Stack<>();
        // push every subtree in inorder traversal to the stack
        // traversal the tree in reverse of inorder traversal
        // right root left
        int val = 0;
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.right;
            }
            cur = stack.pop();
            cur.val = cur.val + val;
            val = cur.val;
            cur = cur.left;
        }
        return root;
    }
}

recursion 版本

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int val = 0;
    public TreeNode bstToGst(TreeNode root) {
        if (root == null) {
            return root;
        }
        if (root.right != null) {
            bstToGst(root.right);
        }
        root.val = val + root.val;
        val = root.val;
        if (root.left != null) {
            bstToGst(root.left);
        }
        return root;
    }
}

2. 1214. Two Sum BSTs
         其中一个pointer放在tree的最小值依次增大，另一个pointer放在另一个tree的最大值，依次减小

/**
 * 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) {
        Stack stack1 = new Stack<>();
        Stack stack2 = new Stack<>();
        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;
            }
            TreeNode t1 = stack1.peek();
            TreeNode t2 = stack2.peek();
            if (t1.val + t2.val == target) {
                return true;
            } else if (t1.val + t2.val < target) {
                stack1.pop();
                root1 = t1.right;
            } else {
                stack2.pop();
                root2 = t2.left;
            }
            
        }
        return false;
    }
}