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目录
104-二叉树的最大深度 Maximum Depth of Binary Tree
问题
深度优先(递归法)
广度优先(队列)方式一
广度优先(队列)方式二
111-二叉树的最小深度 Minimum Depth of Binary Tree
题目
递归

给定二叉树 [3,9,20,null,null,15,7]，3/ \9  20/  \15   7

class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}
}class Solution {public int maxDepth(TreeNode root) {}
}

class Solution {public int maxDepth(TreeNode root) {int depth = 0;if (root != null) {depth = Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;}return depth;}
}

public static void levelTraversal(Node root) {Queue<Node> queue = new LinkedList<>(); //LinkedList是Java中最普通的一个队列queue.offer(root); //add、addLastwhile (!queue.isEmpty()) {Node node = queue.poll();//removeFirstif (node != null) {System.out.print(node.value);queue.offer(node.left);queue.offer(node.right);}}
}

class Solution {public int maxDepth(TreeNode root) {Queue<Pair<TreeNode, Integer>> queue = new LinkedList<>();queue.offer(new Pair<>(root, 0));int depth = 0;while (!queue.isEmpty()) {Pair<TreeNode, Integer> current = queue.poll();depth = Math.max(depth, current.depth); //更新 depthif (current.node != null) {queue.offer(new Pair<>(current.node.left, current.depth + 1));queue.offer(new Pair<>(current.node.right, current.depth + 1));}}return depth;}
}

class Pair<TreeNode, Integer> {TreeNode node;Integer depth;Pair(TreeNode node, Integer depth) {this.node = node;this.depth = depth;}
}

class Solution {public int maxDepth(TreeNode root) {if (root == null) return 0;Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);int depth = 0;while (!queue.isEmpty()) {depth++;int n = queue.size();for (int i = 0; i < n; i++) {TreeNode node = queue.poll();if (node.left != null) queue.offer(node.left);if (node.right != null) queue.offer(node.right);}}return depth;}
}

    3/ \9  20/  \15   7

class Solution {public int minDepth(TreeNode root) {if (root != null) {if (root.left == null && root.right == null) return 1; //这个判断可以省略else if (root.left == null) return minDepth(root.right) + 1;else if (root.right == null) return minDepth(root.left) + 1;else return Math.min(minDepth(root.left), minDepth(root.right)) + 1;} else return 0;}
}