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Leetcode 110 Balanced Binary Tree

发表于 更新于 分类于 Valine：
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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:

img

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

Example 2:

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

Example 3:

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Input: root = []
Output: true

Constraints:

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

自底向上的递归

碰到不平衡的节点就返回高度为-1.

当两个子树都平衡的时候才计算高度

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/**
 * 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 {
    private final static int NotBalanced = -1;
    public boolean isBalanced(TreeNode root) {
        return getDepth(root)[1] != NotBalanced;
    }
    
    private int[] getDepth(TreeNode root)
    {
        if(root == null)
            return new int[]{0,0};
        if(root.left == null && root.right == null)
            return new int[]{1,0};
        int[] leftTree = getDepth(root.left);
        if(leftTree[1] == NotBalanced)
            return new int[]{0, NotBalanced};
        int[] rightTree = getDepth(root.right);
        if(rightTree[1] == NotBalanced)
            return new int[]{0, NotBalanced};
        if(Math.abs(leftTree[0] - rightTree[0]) > 1)
            return new int[]{0, NotBalanced};
        return new int[]{Math.max(leftTree[0], rightTree[0]) + 1, 0};
    }
}