12.判断一棵二叉树是否是平衡二叉树（JavaScript版）
判断一棵二叉树是否是平衡二叉树：
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//二叉平衡搜索树：
//1.根节点的左子树与右子树的高度差不能超过1
//2.这棵二叉树的每个子树都符合第一条
function Node(value){
this.value = value;
this.left = null;
this.right = null;
}
var nodeA = new Node("a");
var nodeB = new Node("b");
var nodeC = new Node("c");
var nodeD = new Node("d");
var nodeE = new Node("e");
var nodeF = new Node("f");
var nodeG = new Node("g");
nodeA.left = nodeB;
nodeA.right = nodeC;
nodeB.left = nodeD;
nodeB.right = nodeE;
nodeC.left = nodeF;
nodeC.right = nodeG;
//判断二叉树是否平衡
function isBalance(root){
if(root == null) return true;
//获取左右的深度
var leftDeep = getDeep(root.left);
var rightDeep = getDeep(root.right);
//若左右深度相差大于1，则不是平衡树
if(Math.abs(leftDeep - rightDeep) > 1){
return false;
}else{
//判断左右子树是否平衡
return isBalance(root.left) && isBalance(root.right);
}
}
//获取树的深度
function getDeep(root){
if(root == null) return 0;
var leftDeep = getDeep(root.left);
var rightDeep = getDeep(root.right);
return Math.max(leftDeep, rightDeep) + 1;//取两个深度中最大的一个，加上自身
}
console.log(isBalance(nodeA));
</script>
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