[Leetcode]110. Balanced Binary Tree

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：求每个节点左右子树的高度，如果  H左树 - H 右树大于 1 或者 H 右树 - H 左树大于 1 ，就返回false ，为此，我另写了个求高度的辅助函数。这么做

我觉得效率有点低。。。但是第一反应就想到的这个。回头看看网上别的大神有没更高效的方法。。

/**
 * 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) {
        int lHeight,rHeight;
        if (root==null)
            return true;                    //如果一开始传进来一个空引用，则返回true
        else{
            lHeight = getHeight(root.left);        //求左树高和右树高
            rHeight = getHeight(root.right);
            if (lHeight-rHeight>1||rHeight-lHeight>1)
                return false;                              //不满足平衡条件，返回false;
            else{
                //满足的话，递归判断左右孩子是否满足条件。左右孩子也可能不满足条件
                //，所以要递归判断。想象下人字形树。
                return isBalanced(root.left)&&isBalanced(root.right);
            }
        }
    }
    public int getHeight(TreeNode root){
        int lHeight=0,rHeight=0;
        if (root==null)
            return 0;
        else {
            lHeight = getHeight(root.left);
            rHeight = getHeight(root.right);
        }
        if (lHeight<rHeight)
            lHeight = rHeight;
        return lHeight+1;
    }
}