二叉树..

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 null
    if (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;
}