04.二叉树的最大深度 （优先掌握递归）111.二叉树的最小深度 （优先掌握递归） 222.完全二叉树的节点个数（优先掌握递归）

04.二叉树的最大深度

/**
 * 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 maxDepth(TreeNode root) {
        
        int a = dfs(root);
        return a;
    }
    int dfs(TreeNode root){
       if (root == null) return 0;
       int left = dfs(root.left);
       int right = dfs(root.right);
       return Math.max(left,right) + 1;
    }
}

111.二叉树的最小深度

/**
 * 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 minDepth(TreeNode root) {
        int a = dfs(root);
        return a;
    }
    public int dfs(TreeNode root){
        if(root == null) return 0;
        int left = dfs(root.left);
        int right = dfs(root.right);
        if(root.left != null && root.right == null){
            return left + 1;
        }
        if(root.left == null && root.right != null){
            return right + 1;
        }
        int result = Math.min(left,right)+1;
        return result;
    }

}

222.完全二叉树的节点个数

采用层次遍历：
时间复杂度O（n）
空间复杂度O（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> queue= new LinkedList<>();
        queue.offer(root);
        int nums = 0;
        while(!queue.isEmpty()){
            int len = queue.size();
            nums += len;
            while(len > 0){
                TreeNode temp = queue.poll();
                if(temp.left != null) queue.add(temp.left);
                if(temp.right != null) queue.add(temp.right);
                len--;
            }
        }
        return nums;
    }
}

如果采用上一种方法，没有使用完全二叉树的条件不太合理。

普通递归法：

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

利用完全二叉树的规律：

时间复杂度（lgn * lgn）

空间复杂度（lgn）

/**
 * 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) {
        int result = f(root);
        return result;
    }
    int f(TreeNode root){
        if(root == null) return 0;
        TreeNode left = root.left;
        TreeNode right = root.right;
        int depthleft = 0,depthright = 0;
        while(left != null){
            left = left.left;
            depthleft++;
        }
        while(right != null){
            right = right.right;
            depthright++;
        }
        if(depthleft == depthright){
            return (2 << depthleft) - 1;
        }
        return f(root.left) + f(root.right) + 1;
    }
}