平衡二叉树

描述

给定一个二叉树, 确定它是高度平衡的。对于这个问题, 一棵高度平衡的二叉树的定义是：一棵二叉树中每个节点的两个子树的深度相差不会超过 1。

样例

给出二叉树 A={3,9,20,#,#,15,7}, B={3,#,20,15,7}

A)  3            B)    3 
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

二叉树 A 是高度平衡的二叉树，但是 B 不是

标签

递归

代码实现

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
class ResultType {
    public boolean isBalanced;
    public int depth;
    public ResultType(boolean isBalanced, int depth) {
        this.isBalanced = isBalanced;
        this.depth = depth;
    }
}
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    public boolean isBalanced(TreeNode root) {
        return helper(root).isBalanced;
    }
    private ResultType helper(TreeNode root) {
        if (root == null) {
            return new ResultType(true, 0);
        }
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        // subtree is not balanced
        if (!left.isBalanced || !right.isBalanced) {
            return new ResultType(false, -1);
        }
        // root is not balanced
        if (Math.abs(left.depth - right.depth) > 1) {
            return new ResultType(false, -1);
        }
        return new ResultType(true, Math.max(left.depth, right.depth) + 1);
    }
}