题解 | #36.判断是不是平衡二叉树#

先求出每一个结点的高度->getHeight(root)

递归地判断是否为平衡二叉树->isBalanced(root)

function IsBalanced_Solution(pRoot)
{
  function getHeight(root){
    if(root == null)  return 0;
    root.height = Math.max(getHeight(root.left), getHeight(root.right)) + 1;
    return root.height;
  }
  function isBalanced(root){
    if(root == null)
      return true;
    if(root.left==null && root.right==null)
      return true;
    if(root.left==null)
      return root.right.height<=1 && isBalanced(root.right);
    if(root.right==null)
      return root.left.height<=1 && isBalanced(root.left);
    
    return Math.abs( root.left.height - root.right.height ) <=1 && isBalanced(root.left) && isBalanced(root.right);
  }
  getHeight(pRoot);
  return isBalanced(pRoot);
  
}
module.exports = {
    IsBalanced_Solution : IsBalanced_Solution
};