平衡二叉树

原题

　　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.

题目大意

　　给定一棵平衡二叉树，判断它是不是高度平衡的。一棵高度平衡的二叉树是左右子树的高度相差不超过1，对其左右子树也是如此。

解题思路

　　递归分治法求解。

代码实现

 

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

 

算法实现类

public class Solution {
    private int minDepth = 0;
    private int maxDepth = 0;

    public boolean isBalanced(TreeNode root) {

        if (root == null) {
            return true;
        }

        int left = depth(root.left);
        int right = depth(root.right);
        if (left - right > 1 || left - right < -1) {
            return false;
        } else {
            return isBalanced(root.left) && isBalanced(root.right);
        }
    }

    /**
     * 求树的高度
     * @param n 树的根结点
     * @return 树的高度
     */
    private int depth(TreeNode n) {
        if (n == null) {
            return 0;
        }  if (n.left == null && n.right == null) {
            return 1;
        } else {
            int left = depth(n.left);
            int right = depth(n.right);
            return 1 + (left > right ? left : right);
        }
    }

}