[代码随想录] 第十二天

144.二叉树的前序遍历[https://leetcode.cn/problems/binary-tree-preorder-traversal/]
思路：栈实现的迭代遍历：出栈记录，右孩子非空右孩子进栈，左孩子非空左孩子进栈。

/**
 * 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<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Stack<TreeNode> sc = new Stack<>();
        sc.push(root);
        TreeNode p;
        while (!sc.isEmpty()) {
            p = sc.pop();
            ans.add(p.val);
            if (p.right != null) {
                sc.push(p.right);
            }
            if (p.left != null) {
                sc.push(p.left);
            }
        }
        return ans;
    }
}

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94.二叉树的中序遍历[https://leetcode.cn/problems/binary-tree-inorder-traversal/submissions/497799353/]
思路：栈实现的迭代遍历。我保持非空一路向左，否则出栈赋值右拐。

/**
 * 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<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> sc = new Stack<>();
        List<Integer> ans = new ArrayList<>();
        TreeNode p = root;
        while(p!=null || !sc.isEmpty()){
            if(p!=null){
            sc.push(p);
            p=p.left;
            }else if(p==null){
                p=sc.pop();
                ans.add(p.val);
                p=p.right;
            }
        }return ans;
}}

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145.二叉树的后序遍历[https://leetcode.cn/problems/binary-tree-postorder-traversal/submissions/497800150/]
思路：二叉树的后序遍历，其实就是前序遍历的思路，当迭代遍历的前序遍历，将非空节点的左右节点进栈的顺序调换并且把结果数组reverse一下就是后序遍历结果。

/**
 * 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<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Stack<TreeNode> sc = new Stack<>();
        sc.push(root);
        TreeNode p;
        while (!sc.isEmpty()) {
            p = sc.pop();
            ans.add(p.val);
            if (p.left != null) {
                sc.push(p.left);
            }
            if (p.right != null) {
                sc.push(p.right);
            }

        }
        Collections.reverse(ans);
        return ans;

    }
}
-----真的理解为什么上述结果为什么就是后序遍历结果并不难，但是我们主动很难去想到，所以我选择死记。-----