feat: 中序遍历

azl397985856 · Dec 29, 2021 · d697340 · d697340
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diff --git a/problems/94.binary-tree-inorder-traversal.md b/problems/94.binary-tree-inorder-traversal.md
@@ -74,41 +74,19 @@ https://leetcode-cn.com/problems/binary-tree-inorder-traversal/
 JavaScript Code：
 
 ```js
-/**
- * @param {TreeNode} root
- * @return {number[]}
- */
 var inorderTraversal = function (root) {
-  // 1. Recursive solution
-  // if (!root) return [];
-  // const left = root.left ? inorderTraversal(root.left) : [];
-  // const right = root.right ? inorderTraversal(root.right) : [];
-  // return left.concat([root.val]).concat(right);
-
-  // 2. iterative solutuon
-  if (!root) return [];
-  const stack = [root];
-  const ret = [];
-  let left = root.left;
-
-  let item = null; // stack 中弹出的当前项
-
-  while (left) {
-    stack.push(left);
-    left = left.left;
-  }
-
-  while ((item = stack.pop())) {
-    ret.push(item.val);
-    let t = item.right;
-
-    while (t) {
-      stack.push(t);
-      t = t.left;
+  const res = [];
+  const stk = [];
+  while (root || stk.length) {
+    while (root) {
+      stk.push(root);
+      root = root.left;
     }
+    root = stk.pop();
+    res.push(root.val);
+    root = root.right;
   }
-
-  return ret;
+  return res;
 };
 ```
 
@@ -143,46 +121,22 @@ public:
 
 Python Code:
 
-```Python
-# Definition for a binary tree node.
-# class TreeNode:
-#     def __init__(self, x):
-#         self.val = x
-#         self.left = None
-#         self.right = None
-
+```py
 class Solution:
     def inorderTraversal(self, root: TreeNode) -> List[int]:
-        """
-        1. 递归法可以一行代码完成，无需讨论；
-        2. 迭代法一般需要通过一个栈保存节点顺序，我们这里直接使用列表
-          - 首先，我要按照中序遍历的顺序存入栈，这边用的逆序，方便从尾部开始处理
-          - 在存入栈时加入一个是否需要深化的参数
-          - 在回头取值时，这个参数应该是否，即直接取值
-          - 简单调整顺序，即可实现前序和后序遍历
-        """
-        # 递归法
-        # if root is None:
-        #     return []
-        # return self.inorderTraversal(root.left)\
-        #     + [root.val]\
-        #     + self.inorderTraversal(root.right)
-        # 迭代法
-        result = []
-        stack = [(1, root)]
-        while stack:
-            go_deeper, node = stack.pop()
-            if node is None:
-                continue
-            if go_deeper:
-                # 左右节点还需继续深化，并且入栈是先右后左
-                stack.append((1, node.right))
-                # 节点自身已遍历，回头可以直接取值
-                stack.append((0, node))
-                stack.append((1, node.left))
-            else:
-                result.append(node.val)
-        return result
+        if not root: return []
+        stack = []
+        ans = []
+        cur = root
+
+        while cur or stack:
+            while cur:
+                stack.append(cur)
+                cur = cur.left
+            cur = stack.pop()
+            ans.append(cur.val)
+            cur = cur.right
+        return ans
 ```
 
 Java Code:
@@ -253,7 +207,6 @@ class Solution {
 
 - [二叉树的遍历](https://github.com/azl397985856/leetcode/blob/master/thinkings/binary-tree-traversal.md)
 
-
 大家对此有何看法，欢迎给我留言，我有时间都会一一查看回答。更多算法套路可以访问我的 LeetCode 题解仓库：https://github.com/azl397985856/leetcode 。 目前已经 37K star 啦。
 大家也可以关注我的公众号《力扣加加》带你啃下算法这块硬骨头。
-![](https://tva1.sinaimg.cn/large/007S8ZIlly1gfcuzagjalj30p00dwabs.jpg)
+![](https://tva1.sinaimg.cn/large/007S8ZIlly1gfcuzagjalj30p00dwabs.jpg)
diff --git a/thinkings/binary-tree-traversal.md b/thinkings/binary-tree-traversal.md
@@ -199,6 +199,10 @@ def MorrisTraversal(root):
 
 **划重点：Morris 是一种可以在 $O(1)$ 空间遍历二叉树的算法。\***
 
+## 相关专题
+
+- [几乎刷完了力扣所有的树题，我发现了这些东西。。。](https://lucifer.ren/blog/2020/11/23/tree/)
+
 ## 相关题目
 
 - [lowest-common-ancestor-of-a-binary-tree](https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/)