[LeetCode] 110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: true

Example 2:

Input: root = [1,2,2,3,3,null,null,4,4]
Output: false

Example 3:

Input: root = []
Output: true

Constraints:

The number of nodes in the tree is in the range [0, 5000].
-104 <= Node.val <= 104

平衡二叉树。

给定一个二叉树，判断它是否是高度平衡的二叉树。

本题中，一棵高度平衡二叉树定义为：

一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。

这个题我提供两种做法，一是通过后序遍历，判断高度是否符合条件。后序遍历的大致思路是先看左子树再看右子树再看当前节点。跑一下这个例子，

Given the following tree [1,2,2,3,3,null,null,4,4]:
       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

后序遍历先跑的是最底层的两个4，两者的高度都是1。再看上面一层的两个3，靠左的3因为有子树的关系所以高度是2，但是靠右的3高度是1，因为没有子树。同理，靠左的2的高度是3，靠右的2的高度是1。当遍历到这一层的时候因为两个2的高度差大于1了，会返回-1。因为树的高度一定是正数所以用-1来代表树的高度差不满足题意。

时间O(n)

空间O(n)

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {boolean}
 4  */
 5 var isBalanced = function (root) {
 6     if (root === null) return true;
 7     return helper(root) !== -1;
 8 };
 9 
10 var helper = function (root) {
11     if (root === null) return 0;
12     let left = helper(root.left);
13     let right = helper(root.right);
14     if (left == -1 || right == -1 || Math.abs(left - right) > 1) {
15         return -1;
16     }
17     return Math.max(left, right) + 1;
18 }

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public boolean isBalanced(TreeNode root) {
12         if (root == null) {
13             return true;
14         }
15         return helper(root) != -1;
16     }
17 
18     private int helper(TreeNode root) {
19         if (root == null) {
20             return 0;
21         }
22         int l = helper(root.left);
23         int r = helper(root.right);
24         if (l == -1 || r == -1 || Math.abs(l - r) > 1) {
25             return -1;
26         }
27         return Math.max(l, r) + 1;
28     }
29 }

第二种做法是类似104题的找树的最大高度的思路。如果左右子树的高度差大于1，则说明这个树不是balanced。时间空间复杂度都是O(n)。这个做法本质上也是后序遍历。

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     boolean balanced = true;
18 
19     public boolean isBalanced(TreeNode root) {
20         helper(root);
21         return balanced;
22     }
23 
24     private int helper(TreeNode root) {
25         if (root == null) {
26             return 0;
27         }
28         int left = helper(root.left);
29         int right = helper(root.right);
30         if (Math.abs(right - left) > 1) {
31             balanced = false;
32         }
33         return Math.max(left, right) + 1;
34     }
35 }

CNoodle

Let the code speak for you.

[LeetCode] 110. Balanced Binary Tree

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