leetcode/110.balanced-binary-tree.java at master · ShawnLi1014/leetcode · GitHub

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/*
 * @lc app=leetcode id=110 lang=java
 *
 * [110] Balanced Binary Tree
 *
 * https://leetcode.com/problems/balanced-binary-tree/description/
 *
 * algorithms
 * Easy (41.10%)
 * Likes:    1290
 * Dislikes: 116
 * Total Accepted:    330.5K
 * Total Submissions: 802K
 * Testcase Example:  '[3,9,20,null,null,15,7]'
 *
 * Given a binary tree, determine if it is height-balanced.
 *
 * For this problem, a height-balanced binary tree is defined as:
 *
 *
 * a binary tree in which the depth of the two subtrees of every node never
 * differ by more than 1.
 *
 *
 * Example 1:
 *
 * Given the following tree [3,9,20,null,null,15,7]:
 *
 *
 * ⁠   3
 * ⁠  / \
 * ⁠ 9  20
 * ⁠   /  \
 * ⁠  15   7
 *
 * Return true.
 *
 * Example 2:
 *
 * Given the following tree [1,2,2,3,3,null,null,4,4]:
 *
 *
 * ⁠      1
 * ⁠     / \
 * ⁠    2   2
 * ⁠   / \
 * ⁠  3   3
 * ⁠ / \
 * ⁠4   4
 *
 *
 * Return false.
 *
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }
        return balancedNode(root) && isBalanced(root.left) && isBalanced(root.right);
    }

    private boolean balancedNode(TreeNode node) {
        if (node == null) {
            return true;
        }
        return Math.abs(depth(node.left) - depth(node.right)) <= 1;
    }

    private int depth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return 1 + Math.max(depth(root.left), depth(root.right));
    }
}
// Time complexity O(n)

class Solution {
    public boolean isBalanced(TreeNode root) {
        return dfsHeight(root) > -1;
    }

    private int dfsHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = dfsHeight(root.left);
        int rightHeight = dfsHeight(root.right);
        if (leftHeight == -1) {
            return -1;
        }
        if (rightHeight == -1) {
            return -1;
        }
        if (Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        return 1 + Math.max(leftHeight, rightHeight);
    }
}
// Time complexity O(n), space complexity O(n)