LeetCode: Binary Tree Depth相关题目合集

最新推荐文章于 2025-09-18 16:18:44 发布

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本文探讨了二叉树的最大深度与最小深度计算方法，并介绍了如何判断一棵二叉树是否为平衡二叉树。通过递归算法实现，考虑了各种边界情况。

Maximum Depth of Binary Tree

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

public class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
        if(root.left == null && root.right == null){
            return 1;
        }
        int left = maxDepth(root.left);
        int right = maxDepth(root.right);
        return Math.max(left, right) + 1;
    }
}

Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

这里与Maximum Depth of Binary Tree基本一样，但是需要考虑极端edge case: root == null这时应返回0.

public class Solution {
    public int minDepth(TreeNode root) {
        if(root == null){
            return 0;
        }
        return depth(root);
    }

    private int depth(TreeNode root){
        if(root == null){
            return Integer.MAX_VALUE;
        }
        if(root.left == null && root.right == null){
            return 1;
        }
        int left = depth(root.left);
        int right = depth(root.right);
        return Math.min(left, right) + 1;
    }
}

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 depth of the two subtrees of every node never differ by more than 1.
这个题目实际上是Maximum Depth of Binary Tree的follow up。直观思路就是判断左子树和右子树的高度差是否大于1.

public class Solution {
    public boolean isBalanced(TreeNode root) {
        return maxDepth(root) != -1;
    }

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