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。