LeetCode之判断高度平衡二叉树

先看问题描述：

/**
 * 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.
 * 
 */

只要左右孩子节点的高度相差不到1，就可以认为这个二叉树是高度平衡二叉树。
判断数的高度，自然要用到递归。

public class BalancedBinaryTree {
    public static boolean isBalanced(TreeNode root) {
        return determine(root) >= 0 ? true : false;
    }

    private static int determine(TreeNode root) {
        if (root == null) {
            return 0;
        } else {
            int leftDepth = determine(root.left);
            int rightDepth = determine(root.right);
            if (leftDepth < 0 || rightDepth < 0|| Math.abs(leftDepth - rightDepth) > 1)
                return -1;
            return Math.max(leftDepth, rightDepth) + 1;
        }
    }
    public static void main(String arg[])
    {
        TreeNode r1 = new TreeNode(1);  
        TreeNode r2 = new TreeNode(2);  
        TreeNode r3 = new TreeNode(3);  
        TreeNode r4 = new TreeNode(4);  
        TreeNode r5 = new TreeNode(5);  
        TreeNode r6 = new TreeNode(6);  

        r1.left = r2;  
        r1.right = r3;  
        r2.left = r4;  
        r2.right = r5;  
        r3.right = r6;  
        if(isBalanced(r1))
        {
            System.out.print("is a height-balanced tree");
        }
        else{
            System.out.print("is not a height-balanced tree");
        }
    }
}

二叉树节点定义：

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

}

上面代码中，determine返回的为正值则表示是高度平衡二叉树。初始化二叉树时简单处理了，比较方便看。