1379 Find a Corresponding Node of a Binary Tree in a Clone of That Tree

Given two binary trees original and cloned and given a reference to a node target in the original tree.

The cloned tree is a copy of the original tree.

Return a reference to the same node in the cloned tree.

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

Follow up: Solve the problem if repeated values on the tree are allowed.

Example 1:

Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.

Example 2:

Input: tree = [7], target =  7
Output: 7

Example 3:

Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4

Example 4:

Input: tree = [1,2,3,4,5,6,7,8,9,10], target = 5
Output: 5

Example 5:

Input: tree = [1,2,null,3], target = 2
Output: 2

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].

• The values of the nodes of the tree are unique.

• target node is a node from the original tree and is not null.

其实这题一看就知道是同时遍历两棵树，可以用bfs队列，或者dfs递归，或者iteration来前中后序遍历。因为遍历模板背的不熟悉，所以选择了用中序stack。T:O(n), S:O(height)

public final TreeNode getTargetCopy(final TreeNode original, final TreeNode cloned, final TreeNode target) {
    if (original == null || cloned == null) {
        return null;
    }

    Stack<TreeNode> stack1 = new Stack<>();
    Stack<TreeNode> stack2 = new Stack<>();
    TreeNode cur1 = original;
    TreeNode cur2 = cloned;
    while (cur1 != null || !stack1.isEmpty()) {
        while (cur1 != null) {
            stack1.push(cur1);
            stack2.push(cur2);
            cur1 = cur1.left;
            cur2 = cur2.left;
        }

        cur1 = stack1.pop();
        cur2 = stack2.pop();

        if (cur1.val == target.val) {
            return cur2;
        }

        cur1 = cur1.right;
        cur2 = cur2.right;
    }

    return null;
}