平衡二叉树

1、空数
2、求root的左右子树的深度，左右子树的深度 差在 1以内

     public boolean IsBalanced_Solution(TreeNode root) {
        if(root == null) return true;
        int l = 0;
        if(root.left != null) {
            l = depth(root.left);
        }
        int r = 0;
        if(root.right != null) {
            r = depth(root.right);
        }
        return Math.abs(l - r) > 1 ? false : true;
    }
    /**
     * 求一棵树的深度
     * @param node
     * @return
     */
    public int depth(TreeNode node) {
        int legnth = 0;
        if(node.left!=null || node.right!=null) {
            legnth ++;
        }
        int l = 0;
        if(node.left != null) {
            l = depth(node.left);
        }
        int r = 0;
        if(node.right != null) {
            r = depth(node.right);
        }
        legnth += (l >= r ? l : r);
        return legnth;
    } 

    public boolean MLR(TreeNode node) {
        if(!helpBalanced(node)){
            return false;
        }
        if(node.left != null) {
            MLR(node.left);
        }
        if(node.right != null) {
            MLR(node.right);
        }
        return true;
    }

    public boolean helpBalanced(TreeNode node) {
        int leftHigh = 0;
        int rightHigh = 0;
        if(node.left != null) {
            leftHigh ++;
            if(node.left.left != null || node.left.right != null) {
                leftHigh ++;
            }
        }
        if(node.right != null) {
            rightHigh ++;
            if(node.right.left != null || node.right.right != null) {
                rightHigh ++;
            }
        }
        return Math.abs(rightHigh - leftHigh) > 1 ? false : true;
    }