前中后序非递归实现通用代码：

LinkedList<TreeNode> stack = new LinkedList<>();TreeNode curr = root; // 代表当前节点
TreeNode pop = null; // 最近一次弹栈的元素
while (curr != null || !stack.isEmpty()) {if (curr != null) {colorPrintln("前: " + curr.val, 31);stack.push(curr); // 压入栈，为了记住回来的路curr = curr.left;} else {TreeNode peek = stack.peek();// 右子树可以不处理, 对中序来说, 要在右子树处理之前打印if (peek.right == null) {colorPrintln("中: " + peek.val, 36);pop = stack.pop();colorPrintln("后: " + pop.val, 34);}// 右子树处理完成, 对中序来说, 无需打印else if (peek.right == pop) {pop = stack.pop();colorPrintln("后: " + pop.val, 34);}// 右子树待处理, 对中序来说, 要在右子树处理之前打印else {colorPrintln("中: " + peek.val, 36);curr = peek.right;}}
}public static void colorPrintln(String origin, int color) {System.out.printf("\033[%dm%s\033[0m%n", color, origin);
}

1. 对称二叉树-力扣 101 题

public boolean isSymmetric(TreeNode root) {return check(root.left, root.right);
}public boolean check(TreeNode left, TreeNode right) {// 若同时为 nullif (left == null && right == null) {return true;}// 若有一个为 null (有上一轮筛选，另一个肯定不为 null)if (left == null || right == null) {return false;}if (left.val != right.val) {return false;}return check(left.left, right.right) && check(left.right, right.left);
}

2. 二叉树最大深度-力扣 104 题

思路：左右根

public int maxDepth(TreeNode node) {if (node == null) {return 0; // 非力扣题目改为返回 -1}int d1 = maxDepth(node.left);int d2 = maxDepth(node.right);return Integer.max(d1, d2) + 1;
}

非递归实现：思路：使用非递归后序遍历, 栈的最大高度即为最大深度

/*思路：1. 使用非递归后序遍历, 栈的最大高度即为最大深度*/
public int maxDepth(TreeNode root) {TreeNode curr = root;LinkedList<TreeNode> stack = new LinkedList<>();int max = 0;TreeNode pop = null;while (curr != null || !stack.isEmpty()) {if (curr != null) {stack.push(curr);int size = stack.size();if (size > max) {max = size;}curr = curr.left;} else {TreeNode peek = stack.peek();if(peek.right == null || peek.right == pop) {pop = stack.pop();} else {curr = peek.right;}}}return max;
}

实现方式三：层序遍历：思路：使用层序遍历, 层数即最大深度

/*思路：1. 使用层序遍历, 层数即最大深度*/
public int maxDepth(TreeNode root) {if(root == null) {return 0;}Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);int level = 0;while (!queue.isEmpty()) {level++;int size = queue.size();for (int i = 0; i < size; i++) {TreeNode node = queue.poll();if (node.left != null) {queue.offer(node.left);}if (node.right != null) {queue.offer(node.right);}}}return level;
}