LeetCode 2313: Minimum Flips in Binary Tree to Get Result Solution

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2313. Minimum Flips in Binary Tree to Get Result

Hard

Dynamic Programming Tree Depth-First Search Binary Tree

Problem Explanation

Explanation:

Given a binary tree where each node has a value of either 0 or 1, we need to find the minimum number of flips required to convert the binary tree to a target binary tree.

To solve this problem, we can perform a depth-first search (DFS) traversal of the tree and simulate flipping the nodes to match the target binary tree. We can recursively check if the current node's value matches the target value, and if not, we need to flip the current node and its children.

Here's the algorithmic idea:

Perform a DFS traversal of the tree.
At each node, check if the current node's value matches the target value.
If not, flip the current node and its children recursively.
Count the number of flips required during the traversal.

Time Complexity: O(N), where N is the number of nodes in the binary tree. Space Complexity: O(H), where H is the height of the binary tree. :

Solution Code

class Solution {
    public int minFlips(TreeNode root, int[] target) {
        return dfs(root, target, 0) ? 0 : -1;
    }
    
    private boolean dfs(TreeNode node, int[] target, int idx) {
        if (node == null) {
            return true;
        }
        
        boolean left = dfs(node.left, target, idx + 1);
        boolean right = dfs(node.right, target, idx + 1);
        
        if (idx < target.length && node.val != target[idx]) {
            node.val = 1 - node.val;
            return false;
        }
        
        return left && right;
    }
}

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