leetcode-tree-110- Balanced Binary Tree

题目:Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

判断一个二叉树是否为平衡二叉树
关键：
1 平衡二叉树的每一个节点的子数深度相差小于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 true

思路

• 高度平衡二叉树是每一个结点的两个子树的深度差不能超过1
• 那么我们肯定需要一个求各个点深度的函数，然后对每个节点的两个子树来比较深度差

所以获取中间值nums[mid]为根节点root，root.left又是[0,mid]的中间值
同理root.right是[mid,nums.lenght]的中间值
然后采用递归即可得到

解法

function maxDepth(root){
    if(!root) return 0
    return 1 + Math.max(maxDepth(root.left),maxDepth(root.right))
}
var isBalanced = function(root) {
    if(!root) return true
    if( Math.abs(maxDepth(root.left)- maxDepth(root.right)) > 1) return false
    return isBalanced(root.left) && isBalanced(root.right)
};

按照上面例子就是
1 queue = [3]
2 node = 3, level = [3], queue = []
3 node.left = 9, node.right = 20, queue = [9,20]

下一轮循环
1 queue = [9,20]
2 node = 9,level = [9],queue = [20]
3 node = 20,level = [9,20], queue = []
4 node.left = 15, node.right = 7,queue = [17,7]

再一次下一轮。。。