代码随想录算法训练营20期|第十四天|● 理论基础● 递归遍历 ● 迭代遍历● 统一迭代

•  理论基础

二叉树的种类：

满二叉树：

完全二叉树：compete tree，只有最底层靠右的部分有节点缺失

二叉搜索树：所有的左孩子都小于根节点，左右的右孩子都大于根节点

平衡二叉搜索树：AVL

•  递归遍历  

前序遍历

144. 二叉树的前序遍历

/**
 * 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 List preorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        traverse(root, res);
        return res;
    }

    private void traverse(TreeNode root, List res) {
        if (root == null) return;
        res.add(root.val);
        traverse(root.left, res);
        traverse(root.right, res);
    }
}

中序遍历

/**
 * 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 List inorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        traverse(root, res);
        return res;
    }

    private void traverse(TreeNode root, List res) {
        if (root == null) return;
        traverse(root.left, res);
        res.add(root.val);
        traverse(root.right, res);
    }
}

后续遍历

/**
 * 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 List postorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        traverse(root, res);
        return res;
    }

    private void traverse(TreeNode root, List res) {
        if (root == null) return;
        traverse(root.left, res);
        traverse(root.right, res);
        res.add(root.val);
    }
}

•  迭代遍历

前序遍历，用Stack，先root，再右孩子，再左孩子

/**
 * 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 List preorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if (root == null) return res;

        Stack stack = new Stack<>();
        stack.add(root);

        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) stack.add(cur.right);
            if (cur.left != null) stack.add(cur.left);
        }
        return res;
    }
}

中序遍历

要有个指针

/**
 * 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 List inorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if (root == null) return res;
        Stack stack = new Stack<>();
        TreeNode cur = root;

        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                stack.add(cur);
                cur = cur.left;
            } else {
                cur = stack.pop();
                res.add(cur.val);
                cur = cur.right;
            }
        }
        return res;

    }
}

后序遍历

Stack, 按正常的层序遍历来traverse，中左右，出栈顺序是中->右->左，然后把结果反转，左-》右-》中

/**
 * 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 List postorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if (root == null) return res;
        Stack stack = new Stack<>();
        stack.add(root);

        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.left != null) stack.add(cur.left);
            if (cur.right != null) stack.add(cur.right);
        }
        Collections.reverse(res);
        return res;
    }
}

•  统一迭代