day16 打卡104.二叉树的最大深度 559. N 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

day16 打卡104.二叉树的最大深度 559. N 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

104题目链接

1.递归法

class Solution {
    public int maxDepth(TreeNode root) {
        return maxDepth(root, 0);
    }

    public int maxDepth(TreeNode cur, int depth) {
        if (cur == null) return depth;

        depth++;
        int left = maxDepth(cur.left, depth);
        int right = maxDepth(cur.right, depth);
        return left > right ? left : right;
    }
}

2.迭代法,即层次遍历

class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> que = new LinkedList<>();
        que.offer(root);
        int size = 1;
        int depth = 0;
        while (!que.isEmpty()) {
            size = que.size();
            depth++;
            while (size>0) {
                TreeNode cur = que.poll();
                if (cur.left != null) que.offer(cur.left);
                if (cur.right != null) que.offer(cur.right);
                size--;
            }
        }
        return depth;
    }
}

559. N 叉树的最大深度

559题目链接

递归法,与104题目一样的思路

class Solution {
    public int maxDepth(Node root) {
        if (root == null) return 0;

        int depth = 0;
        if (root.children != null) {
            for (Node node: root.children) {
                depth = Math.max(depth, maxDepth(node));
            }
        }
        return depth + 1;
    }
}

111.二叉树的最小深度

111题目链接

1.递归法

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

        int leftDepth = minDepth(root.left);
        int rightDepth = minDepth(root.right);

        if (root.left == null && root.right != null) {
            return rightDepth + 1;
        }
        if (root.left != null && root.right == null) {
            return leftDepth + 1;
        }
        return Math.min(leftDepth, rightDepth) + 1;
    }
}

2.迭代法,即层次遍历

class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> que = new LinkedList<>();
        que.offer(root);
        int size = 1;
        int depth = 0;
        while (!que.isEmpty()) {
            size = que.size();
            depth++;
            while (size>0) {
                TreeNode cur = que.poll();
                if (cur.left != null) que.offer(cur.left);
                if (cur.right != null) que.offer(cur.right);

                if (cur.left == null && cur.right == null) return depth;

                size--;
            }
        }
        return depth;
    }
}

222.完全二叉树的节点个数

222题目链接

1.递归法

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

2.迭代法,即层次遍历

class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) return 0;
        Queue<TreeNode> que = new LinkedList<>();
        que.offer(root);
        int size = 1;
        int count = 0;
        while (!que.isEmpty()) {
            size = que.size();
            while (size>0) {
                TreeNode cur = que.poll();
                count++;
                if (cur.left != null) que.offer(cur.left);
                if (cur.right != null) que.offer(cur.right);
                size--;
            }
        }
        return count;
    }
}

3.针对完全二叉树的解法,满二叉树的结点数为：2^depth - 1

class Solution {
    public int countNodes(TreeNode node) {
        if (node == null) return 0;
        TreeNode left = node.left;
        TreeNode right = node.right;
        int leftCount = 0;
        int rightCount = 0;
        while (left != null) {
            left = left.left;
            leftCount++;
        }
        while (right != null) {
            right = right.right;
            rightCount++;
        }
        if (leftCount == rightCount) {
            // 说明是满二叉树
            return ( 2 << leftCount) - 1;
        }
        return countNodes(node.left) + countNodes(node.right) + 1;
    }
}

参考资料

代码随想录