110. Balanced Binary Tree

    /*
    * 110. Balanced Binary Tree
    * 2016-5-17 by Mingyang
    * This improved algorithm works by checking the height of each subtree as we recurse
    * down from the root. On each node, we recursively get the heights of the left and right
    * subtrees through the checkHeight method. If the subtree is balanced, then check-
    * Height will return the actual height of the subtree. If the subtree is not balanced, then
    * checkHeight will return -1. We will immediately break and return -1 from the current call.
    * 自己刚开始做的时候，在主函数里面吧后面两个条件一起写在一个语句里面了，殊不知base case里面并没有false的statement
    */
   /*
    *Wrong one:
    *return isBalanced(root.left)&&isBalanced(root.right)&&(Math.abs(depth(root.left)-depth(root.right))<=1);
    *太傻叉了这种写法，基本的递归都不会，base case都没写好，如何写递归？
    *以后写递归以前，一定要知道哪些基本case，都没有return false的情况，就会进入死循环。
    */
    public boolean isBalanced(TreeNode root) {
        if (root == null)
            return true;
        if (Math.abs(depth(root.left) - depth(root.right)) > 1)
            return false;
        return isBalanced(root.left) && isBalanced(root.right);
    }
    private int depth(TreeNode root) {
        if (root == null)
            return 0;
        return Math.max(depth(root.left), depth(root.right)) + 1;
    }