110. 平衡二叉树

110. 平衡二叉树

原题

/**
 * 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 {
    //用来判断是否满足平衡二叉树
    boolean flag = true;
    public boolean isBalanced(TreeNode root) {
        dfs(root);
        return flag;
    }

    public int dfs(TreeNode root){
        if(root==null){
            return 0;
        }
        //深度优先遍历左子树
        int leftHeight = dfs(root.left);
        //深度优先遍历右子树
        int rightHeight = dfs(root.right);
        //不满足平衡二叉树
        if(Math.abs(leftHeight-rightHeight)>1){
            flag = false;
        }
        //返回树的高度(这里没有包含当前结点)
        return Math.max(leftHeight,rightHeight)+1;
    }
}