Question

面试题 55 - II. 平衡二叉树

题目描述

输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

 

示例 1:

给定二叉树 [3,9,20,null,null,15,7]

  3
   / \
  9  20
  /  \
   15   7

返回 true 。

示例 2:

给定二叉树 [1,2,2,3,3,null,null,4,4]

     1
    / \
   2   2
  / \
   3   3
  / \
 4   4

返回 false 。

 

限制：

• 0 <= 树的结点个数 <= 10000

注意：本题与主站 110 题相同：https://leetcode.cn/problems/balanced-binary-tree/

 

解法

方法一：递归

我们设计一个递归函数 $dfs(root)$，函数返回值为 $root$ 节点的深度，如果 $root$ 节点不平衡，返回值为 $-1$。

函数 $dfs(root)$ 的递归过程如下：

• 如果 $root$ 为空，返回 $0$。
• 递归计算左右子树的深度，记为 $l$ 和 $r$。
• 如果 $l$ 或 $r$ 为 $-1$，或者 $l$ 和 $r$ 的差的绝对值大于 $1$，返回 $-1$。
• 否则，返回 $max(l, r) + 1$。

如果 $dfs(root)$ 返回值不为 $-1$，则说明 $root$ 节点平衡，返回 true，否则返回 false。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点数。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def isBalanced(self, root: TreeNode) -> bool:
    def dfs(root):
      if root is None:
        return 0
      l, r = dfs(root.left), dfs(root.right)
      if l == -1 or r == -1 or abs(l - r) > 1:
        return -1
      return 1 + max(l, r)

    return dfs(root) != -1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  public boolean isBalanced(TreeNode root) {
    return dfs(root) != -1;
  }

  private int dfs(TreeNode root) {
    if (root == null) {
      return 0;
    }
    int l = dfs(root.left);
    int r = dfs(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(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
  bool isBalanced(TreeNode* root) {
    function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
      if (!root) {
        return 0;
      }
      int l = dfs(root->left);
      int r = dfs(root->right);
      if (l == -1 || r == -1 || abs(l - r) > 1) {
        return -1;
      }
      return 1 + max(l, r);
    };
    return dfs(root) != -1;
  }
};

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

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

// 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 {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
    match root {
      None => 0,
      Some(node) => {
        let node = node.borrow();
        1 + Self::dfs(&node.left).max(Self::dfs(&node.right))
      }
    }
  }
  pub fn is_balanced(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
    match root {
      None => true,
      Some(node) => {
        let mut node = node.borrow_mut();
        let a = 10;
        (Self::dfs(&node.left) - Self::dfs(&node.right)).abs() <= 1 &&
          Self::is_balanced(node.left.take()) &&
          Self::is_balanced(node.right.take())
      }
    }
  }
}

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  public bool IsBalanced(TreeNode root) {
    return dfs(root) != -1;
  }

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

方法二

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def isBalanced(self, root: TreeNode) -> bool:
    def dfs(root):
      if root is None:
        return (True, 0)
      l, ld = dfs(root.left)
      r, rd = dfs(root.right)
      d = max(ld, rd) + 1
      if l and r and abs(ld - rd) <= 1:
        return (True, d)
      return (False, d)

    return dfs(root)[0]