题目链接

后序遍历高度，高度判断是否平衡
前序遍历深度

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null){
            return true;
        }
        return getHeight(root) == -1 ? false : true;
    }

	// -1表示已经不是平衡二叉树了，否则返回值是以该节点为根节点树的高度
    public int getHeight(TreeNode root){
        if(root == null){
            return 0;
        }
        int lh = getHeight(root.left);
        if(lh == -1){
            return -1;
        }
        int rh = getHeight(root.right);
        if(rh == -1){
            return -1;
        }
        // 判断当前传入节点为根节点的二叉树是否为平衡二叉树：求左子树与右子树高度差值绝对值 <= 则为平衡二叉树
        int res;
        if(Math.abs(lh - rh) > 1){
            res = -1;
        }else{
            res = 1 + Math.max(lh,rh);
        }
        return res;
    }
}