Question

110. 平衡二叉树

English Version

题目描述

给定一个二叉树，判断它是否是 平衡二叉树  

 

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：true

示例 2：

输入：root = [1,2,2,3,3,null,null,4,4]
输出：false

示例 3：

输入：root = []
输出：true

 

提示：

• 树中的节点数在范围 [0, 5000] 内
• -104 <= Node.val <= 104

解法

方法一：自底向上的递归

定义函数 $height(root)$ 计算二叉树的高度，处理逻辑如下：

• 如果二叉树 $root$ 为空，返回 $0$。
• 否则，递归计算左右子树的高度，分别为 $l$ 和 $r$。如果 $l$ 或 $r$ 为 $-1$，或者 $l$ 和 $r$ 的差的绝对值大于 $1$，则返回 $-1$，否则返回 $max(l, r) + 1$。

那么，如果函数 $height(root)$ 返回的是 $-1$，则说明二叉树 $root$ 不是平衡二叉树，否则是平衡二叉树。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def isBalanced(self, root: Optional[TreeNode]) -> bool:
    def height(root):
      if root is None:
        return 0
      l, r = height(root.left), height(root.right)
      if l == -1 or r == -1 or abs(l - r) > 1:
        return -1
      return 1 + max(l, r)

    return height(root) >= 0

/**
 * 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) {
    return height(root) >= 0;
  }

  private int height(TreeNode root) {
    if (root == null) {
      return 0;
    }
    int l = height(root.left);
    int r = height(root.right);
    if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
      return -1;
    }
    return 1 + Math.max(l, r);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  bool isBalanced(TreeNode* root) {
    function<int(TreeNode*)> height = [&](TreeNode* root) {
      if (!root) {
        return 0;
      }
      int l = height(root->left);
      int r = height(root->right);
      if (l == -1 || r == -1 || abs(l - r) > 1) {
        return -1;
      }
      return 1 + max(l, r);
    };
    return height(root) >= 0;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isBalanced(root *TreeNode) bool {
  var height func(*TreeNode) int
  height = func(root *TreeNode) int {
    if root == nil {
      return 0
    }
    l, r := height(root.Left), height(root.Right)
    if l == -1 || r == -1 || abs(l-r) > 1 {
      return -1
    }
    if l > r {
      return 1 + l
    }
    return 1 + r
  }
  return height(root) >= 0
}

func abs(x int) int {
  if x < 0 {
    return -x
  }
  return x
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function isBalanced(root: TreeNode | null): boolean {
  const dfs = (root: TreeNode | null) => {
    if (root == null) {
      return 0;
    }
    const left = dfs(root.left);
    const right = dfs(root.right);
    if (left === -1 || right === -1 || Math.abs(left - right) > 1) {
      return -1;
    }
    return 1 + Math.max(left, right);
  };
  return dfs(root) > -1;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  pub fn is_balanced(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    Self::dfs(&root) > -1
  }

  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
    if root.is_none() {
      return 0;
    }
    let node = root.as_ref().unwrap().borrow();
    let left = Self::dfs(&node.left);
    let right = Self::dfs(&node.right);
    if left == -1 || right == -1 || (left - right).abs() > 1 {
      return -1;
    }
    1 + left.max(right)
  }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function (root) {
  const height = root => {
    if (!root) {
      return 0;
    }
    const l = height(root.left);
    const r = height(root.right);
    if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
      return -1;
    }
    return 1 + Math.max(l, r);
  };
  return height(root) >= 0;
};