代码随想录训练营第三十五期|第天16|二叉树part03|104.二叉树的最大深度 ● 111.二叉树的最小深度● 222.完全二叉树的节点个数

104. 二叉树的最大深度 - 力扣（LeetCode）

递归，可以前序遍历，也可以后序遍历

前序遍历是backtracking

下面是后序遍历的代码：

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

层序遍历，到最后一层, 记录遍历了多少层。需要遍历到最后一层

/*** 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) {if (root == null) return 0;Queue<TreeNode> queue = new LinkedList<>();queue.add(root);int level = 0;while (!queue.isEmpty()) {int size = queue.size();level++;for (int i = 0; i < size; i++) {TreeNode cur = queue.poll();if (cur.left != null) queue.add(cur.left);if (cur.right != null) queue.add(cur.right);}}return level;}
}

111. 二叉树的最小深度 - 力扣（LeetCode）

递归：当一边是空的时候，返回另外一边

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

迭代：

当当前的node的左右孩子都为null的时候，就可以返回level了

/*** 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) {if (root == null) return 0;Queue<TreeNode> queue = new LinkedList<>();queue.add(root);int level = 0;while (!queue.isEmpty()) {int size = queue.size();level++;for (int i = 0; i < size; i++) {TreeNode cur = queue.poll();if (cur.left == null && cur.right == null) return level;if (cur.left != null) queue.add(cur.left);if (cur.right != null) queue.add(cur.right);}}return level;}
}

222. 完全二叉树的节点个数 - 力扣（LeetCode）

递归：

/*** 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 left = countNodes(root.left);int right = countNodes(root.right);return left + right + 1;}
}

迭代：层序遍历，每取出一个node，count + 1

/*** 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.add(root);int count = 0;while (!queue.isEmpty()) {int size = queue.size();level++;for (int i = 0; i < size; i++) {TreeNode cur = queue.poll();count++;if (cur.left != null) queue.add(cur.left);if (cur.right != null) queue.add(cur.right);}}return count;}
}