LeetCode

题目：

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.

思路：

1) 计算结点的高度，但这个有重复计算

package tree;

public class BalancedBinaryTree {
    
    public boolean isBalanced(TreeNode root) {
        if (root == null) return true;
        int leftHeight = height(root.left);
        int rightHeight = height(root.right);
        return Math.abs(leftHeight - rightHeight) <= 1 &&
                isBalanced(root.left) && isBalanced(root.right);
    }
    
    private int height(TreeNode root) {
        if (root == null) return 0;
        return Math.max(1 + height(root.left), 1 + height(root.right));
    }
    
}

2) 不进行重复计算，但需要new 对象，反而也花时间

package tree;

public class BalancedBinaryTree {
    
    class Entry{
        public int height;
        public boolean balanced;
    }
    
    public boolean isBalanced(TreeNode root) {
        return checkTree(root).balanced;
    }
    
    private Entry checkTree(TreeNode root) {
        Entry entry = new Entry();
        if (root == null) {
            entry.height = 0;
            entry.balanced = true;
        } else {
            Entry left = checkTree(root.left);
            Entry right = checkTree(root.right);
            entry.height = Math.max(left.height, right.height) + 1;
            entry.balanced = Math.abs(left.height - right.height) <= 1 && left.balanced && right.balanced;
        }
        return entry;
    }
    
}