Leetcode 222. Count Complete Tree Nodes

最新推荐文章于 2024-01-20 11:36:34 发布

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#complete tree #dfs #bfs #binary search

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本文对比三种计数二叉树节点的方法：迭代队列法（时间复杂度n，空间复杂n）、递归前序遍历（时间复杂n，空间复杂h），以及利用二分搜索结合递归的优化方案（初始空间复杂h，优化后为1）。详细介绍了每种方法的实现和空间优化策略。

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方法1: bsf-two queues-iteration。时间复杂n，空间复杂n

/**
 * 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 {
    public int countNodes(TreeNode root) {
        if(root == null) return 0;
        Queue<TreeNode> currLevel = new LinkedList<>();
        Queue<TreeNode> nextLevel = new LinkedList<>();
        nextLevel.offer(root);
        int count = 0;
        while(!nextLevel.isEmpty()){
            currLevel = new LinkedList<>(nextLevel);
            nextLevel.clear();
            count += currLevel.size();
            for(TreeNode node : currLevel){
                if(node.left != null) nextLevel.offer(node.left);
                if(node.right != null) nextLevel.offer(node.right);
            }  
        }
        return count;
    }
}

方法2: dfs-preorder-recursion.时间复杂n，空间复杂h。

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

方法3: binary search + recursion。时间复杂logn的平方，空间复杂h。这个方法充分利用了complete tree这个特质，我们只需要计算最后一层的node数量就好。

/**
 * 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 {
    public int countNodes(TreeNode root) {
        if(root == null) return 0;
        int leftDepth = getDepth(root.left);
        int rightDepth = getDepth(root.right);
        if(leftDepth == rightDepth){
            return 1 + (int)Math.pow(2,leftDepth) - 1 + countNodes(root.right);
        }else{
            return 1 + countNodes(root.left) + (int)Math.pow(2,rightDepth) - 1;
        }
    }
    
    public int getDepth(TreeNode root){
        if(root == null) return 0;
        int depth = 1;
        while(root.left != null){
            depth++;
            root = root.left;
        }
        return depth;
    }
}