Flip Binary Tree To Match Preorder Traversal

Given a binary tree withNnodes, each node has a different value from {1, ..., N}.

A node in this binary tree can be flipped by swapping the left child and the right child of that node.

Consider the sequence of Nvalues reported by a preorder traversal starting from the root. Call such a sequence ofNvalues the voyage of the tree.

(Recall that apreorder traversal of a node means we report the current node's value, then preorder-traverse the left child, then preorder-traverse the right child.)

Our goal is to flip the least number of nodes in the tree so that the voyage of the tree matches thevoyagewe are given.

If we can do so, then return a list of the values of all nodes flipped. You may return the answer in any order.

If we cannot do so, then return the list[-1].

Example

Example 1:

Input: 
root = [1,2], voyage = [2,1]
Output: 
[-1]

Example 2:

Input: 
root = [1,2,3], voyage = [1,3,2]
Output: 
[1]

Example 3:

Input: 
root = [1,2,3], voyage = [1,2,3]
Output: 
[]

Note

逻辑就是进行前序遍历树，同时遍历数组，在相等的遇到不相等的左孩子，就反转，然后递归，如果遇到不相等的情况，就直接返回false

实现上可以递归分治，也可以栈做前序遍历

Code

public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) {
    List<Integer> res = new ArrayList<>();
    int i = 0;
    Stack<TreeNode> s = new Stack<>();
    s.push(root);
    while (!s.isEmpty()) {
        TreeNode curr = s.pop();
        if (curr == null) { continue; }
        if (curr.val != voyage[i++]) { return Arrays.asList(-1); }
        if (curr.right != null && curr.right.val == voyage[i]) {
            if (curr.left != null) { // 针对1-null-2，【1，2】的情况
                res.add(curr.val);
            }
            s.push(curr.left);
            s.push(curr.right);
        } else {
            s.push(curr.right);
            s.push(curr.left);
        }
    }

    return res;
}

List<Integer> res = new ArrayList<>();
int i = 0;
public List<Integer> flipMatchVoyage(TreeNode root, int[] voyage) {
    if (dfs(root, voyage)) {
        return res;
    }  
    return Arrays.asList(-1);
}

private boolean dfs(TreeNode root, int[] v) {
    if (root == null) {
        return true;
    }
    if (root.val != v[i++]) {
        return false;
    }
    if (root.left != null && root.left.val != v[i]) {
        res.add(root.val);
        return dfs(root.right, v) && dfs(root.left, v);
    } else {
        return dfs(root.left, v) && dfs(root.right, v);
    }
}