feat: 算法-二叉树的中序遍历

feat: 算法-二叉树的中序遍历

Author: liyonglu

前端算法练习/04：树、二叉树、二次搜索树/24-binary-tree-inorder-traversal.js

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+// https://leetcode-cn.com/problems/binary-tree-inorder-traversal/
+// 二叉树的中序遍历
+/**
+ * 二叉树的前中后，是根据根节点的遍历位置进行控制的
+ *左-根-右
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+// 1. 递归 O(n)  O(n)
+var inorderTraversal = function (root) {
+  const res = [];
+  const inorder = (root) => {
+    if (!root) {
+      return;
+    }
+    inorder(root.left);
+    res.push(root.val);
+    inorder(root.right);
+  };
+  inorder(root);
+  return res;
+};
+
+// 2. 迭代 O(n)  O(n), 两种方式是等价的，区别在于递归的时候隐式地维护了一个栈，而我们在迭代的时候需要显式地将这个栈模拟出来，其他都相同，
+var inorderTraversal = function (root) {
+  const res = [];
+  const stack = [];
+  while (root || stack.length) {
+    while (root) {
+      stack.push(root);
+      root = root.left;
+    }
+    root = stack.pop();
+    res.push(root.val);
+    root = root.right;
+  }
+  return res;
+};
+
+var inorderTraversal = function (root) {
+  const stack = [];
+  const res = [];
+
+  while (root || stack.length) {
+    if (root) {
+      stack.push(root);
+      root = root.left;
+    } else {
+      root = stack.pop();
+      res.push(root.val);
+      root = root.right;
+    }
+  }
+
+  return res;
+};
+
+//  中序遍历：左中右
+//  压栈顺序：右中左
+
+var inorderTraversal = function (root, res = []) {
+  const stack = [];
+  if (root) stack.push(root);
+  while (stack.length) {
+    const node = stack.pop();
+    if (!node) {
+      res.push(stack.pop().val);
+      continue;
+    }
+    if (node.right) stack.push(node.right); // 右
+    stack.push(node); // 中
+    stack.push(null);
+    if (node.left) stack.push(node.left); // 左
+  }
+  return res;
+};