题解 | #判断是不是平衡二叉树#TOP36

思路:
1.平衡二叉树:左子树和右子树的高度相差不超过1
2.先计算当前左子树的高度和右子树的高度是否相差小于等于1，如果是那说明是平衡的，需要再进行判断当前左子树和右子树，他们的左右子树是否满足平衡条件，不断递归下去
3.递归条件isBanalced(root.left) && isBanalced(root.right) && [Math.abs(getHeight(root.right),getHeight(root.left)) <= 1]

public class Solution {
    boolean isBanalced = true;
    public boolean IsBalanced_Solution(TreeNode root) {
        if(root == null){
            return true;
        }
        //getHeight(root);
        //return isBanalced;
        return isBanalced(root);
    }
    private boolean isBanalced(TreeNode root){
        if(root == null){
            return true;
        }
        int left = getHeight(root.left);
        int right = getHeight(root.right);
        if(Math.abs(right - left) > 1){
            return false;
        }
        return isBanalced(root.left) && isBanalced(root.right);
    }
    
    private int getHeight(TreeNode node){
        if(node == null){
            return 0;
        }
        int left = getHeight(node.left);
        int right = getHeight(node.right);
        if(Math.abs(right - left) > 1){
            isBanalced = false;
        }
        return Math.max(left, right)+1;
    }
}