题目

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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7
Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4
Return false.

思路：

1. 所谓平衡二叉树，如果每个节点的左右两子树深度差绝对值不超过1；
2. 递归；以一个最普通的二叉树思考，该节点的|左子树深度 -左子树深度|<=1，且它的左子树和右子树也满足-->递归
3. 如何获取该节点的深度，这里又是另外一个递归，以一个最普通的二叉树思考：
   leftHight >= rightHight ? leftHight + 1 : rightHight + 1;

代码

public class BalancedBinaryTree {
    public boolean isBalanced(TreeNode root) {
        if(root!=null){
            int leftHight = TreeDepth(root.left);
            int rightHight = TreeDepth(root.right);
            if(leftHight -rightHight < - 1 || leftHight -rightHight >1){
                return false;
            }
            return isBalanced(root.left) && isBalanced(root.right);
        }
        return true;
    }

    public int TreeDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHight = TreeDepth(root.left);
        int rightHight = TreeDepth(root.right);
        return leftHight >= rightHight ? leftHight + 1 : rightHight + 1;
    }
}