不讲武德的解法

java 实现

class Solution {
    public int countNodes(TreeNode root) {
        if(root == null) return 0;
        return countNodes(root.left) + countNodes(root.right) + 1;
    }
}

根据完全二叉树和满二叉树的性质做

class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        int leftDepth = getDepth(root.left);
        int rightDepth = getDepth(root.right);
        
        if (leftDepth == rightDepth) {
            // 左子树是满二叉树，节点数为 2^leftDepth - 1，加上根节点和右子树
            return (1 << leftDepth) + countNodes(root.right);
        } else {
            // 右子树是满二叉树，节点数为 2^rightDepth - 1，加上根节点和左子树
            return (1 << rightDepth) + countNodes(root.left);
        }
    }
    
    private int getDepth(TreeNode node) {
        int depth = 0;
        while (node != null) {
            depth++;
            node = node.left; // 完全二叉树的最左路径
        }
        return depth;
    }
}