1261. Find Elements in a Contaminated Binary Tree

QUESTION:

Given a binary tree with the following rules:

root.val == 0 If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1 If treeNode.val == x and treeNode.right != null, then treeNode.right.val == 2 * x + 2 Now the binary tree is contaminated, which means all treeNode.val have been changed to -1.

You need to first recover the binary tree and then implement the FindElements class:

FindElements(TreeNode* root) Initializes the object with a contamined binary tree, you need to recover it first. bool find(int target) Return if the target value exists in the recovered binary tree.

Example 1:

Input [“FindElements”,”find”,”find”] [[[-1,null,-1]],[1],[2]] Output [null,false,true] Explanation FindElements findElements = new FindElements([-1,null,-1]); findElements.find(1); // return False findElements.find(2); // return True Example 2:

Input [“FindElements”,”find”,”find”,”find”] [[[-1,-1,-1,-1,-1]],[1],[3],[5]] Output [null,true,true,false] Explanation FindElements findElements = new FindElements([-1,-1,-1,-1,-1]); findElements.find(1); // return True findElements.find(3); // return True findElements.find(5); // return False Example 3:

Input [“FindElements”,”find”,”find”,”find”,”find”] [[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]] Output [null,true,false,false,true] Explanation FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]); findElements.find(2); // return True findElements.find(3); // return False findElements.find(4); // return False findElements.find(5); // return True

Constraints:

TreeNode.val == -1 The height of the binary tree is less than or equal to 20 The total number of nodes is between [1, 10^4] Total calls of find() is between [1, 10^4] 0 <= target <= 10^6

EXPLANATION:

其实这道题目看起来是一个中等的题目，但是其实是两个easy的题目加起来的。一个是将被污染的二叉树进行恢复，恢复完成后进行查找。而这两步其实都是考的二叉树的遍历。无论你使用前序，中序，后续遍历其实都可以。
第一步：将污染的二叉树进行恢复。那么肯定是采用递归的方式，恢复的话需要知道parent的值，同时需要知道自己是在左树上还是在右树上，那既然需要知道这些信息，那么我们就将这些信息传递过去。递归的时候将每次的root.val和是否是左树进行传递。递归中进行计算当前root的值，同时再向下传递，递归结束就可以。
第二步：二叉树查找就更不用说了，前中后序随便选择一个，查看左右子树是否找到，进行或运算，只要其中一个找到了就是可以的。

SOLUTION:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class FindElements {
        TreeNode root;

        public FindElements(TreeNode root) {
            this.root = root;
            if (root != null) root.val = 0;
            if (root.left != null)
                FindElementsHelper(root.left, 0, true);
            if (root.right != null)
                FindElementsHelper(root.right, 0, false);
        }

        private void FindElementsHelper(TreeNode root,int value,boolean left){
            if(left) root.val = 2*value+1;
            else root.val = 2*value+2;
            if (root.left != null)
                FindElementsHelper(root.left, root.val, true);
            if (root.right != null)
                FindElementsHelper(root.right, root.val, false);
        }

        public boolean find(int target) {
            return findHelper(target,root);
        }

        public boolean findHelper(int target,TreeNode node){
            if(node.val == target) return true;
            boolean left= false;
            boolean right = false;
            if(node.left!=null) left = findHelper(target,node.left);
            if(node.right!=null) right = findHelper(target,node.right);
            return left|right;
        }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * FindElements obj = new FindElements(root);
 * boolean param_1 = obj.find(target);
 */
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