Leetcode 993. Cousins in Binary Tree

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In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Note:

1. The number of nodes in the tree will be between 2 and 100.
2. Each node has a unique integer value from 1 to 100.

public boolean isCousins(TreeNode root, int x, int y) {
        Map<Integer, TreeNode> parent = new HashMap<>();
        //traversal by layer
        Queue<TreeNode> queue = new LinkedList<>();
        parent.put(root.val, null);
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode peek = queue.poll();
            System.out.println(peek.val + " ");
            if (peek.left != null) {
                queue.offer(peek.left);
                parent.put(peek.left.val, peek);
            }
            if (peek.right != null) {
                queue.offer(peek.right);
                parent.put(peek.right.val, peek);
            }
        }
        TreeNode parentX = parent.get(x);
        TreeNode parentY = parent.get(y);
        int depthX = 0, depthY = 0;
        int val = x;
        while (parent.get(val) != null) {
            depthX++;
            val = parent.get(val).val;
        }

        val = y;
        while (parent.get(val) != null) {
            depthY++;
            val = parent.get(val).val;
        }

        return (parentX != parentY) && (depthX == depthY);
    }