102. 二叉树的层序遍历

102. 二叉树的层序遍历.js

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+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number[][]}
+*/
+// 广度优先通过队列处理 【深度优先用栈】
+// 方法一：BFS通用模版 广度优先：迭代法 时间O(n)、空间O(n)
+var levelOrder = function (root) {
+  let result = [], list = [root]
+  while (list.length) {
+    let len = list.length, level = [], cur
+    while (len--) {
+      cur = list.shift()
+      if (cur !== null) {
+        level.push(cur.val)
+        cur.left !== null && list.push(cur.left)
+        cur.right !== null && list.push(cur.right)
+      }
+    }
+    level.length && result.push(level)
+  }
+  return result
+}
+
+// 方法二 标记法，偏技巧
+var levelOrder = function (root) {
+  if (!root) return []
+  let result = [], level = [], list = [root, null]
+  while (list.length > 0) {
+    const cur = list.shift()
+    if (cur) {
+      level.push(cur.val)
+      cur.left && list.push(cur.left)
+      cur.right && list.push(cur.right)
+    } else { // 一层已经遍历完了
+      result.push(level)
+      level = []
+      list.length > 0 && list.push(null)
+    }
+  }
+  return result
+}
+
+// 方法三 递归 时间O(n),空间O(h) h为树高度
+var levelOrder = function (root) {
+  if (!root) return []
+  let result = []
+  function dfs(cur, level) {
+    if (cur != null) {
+      !result[level] && (result[level] = [])
+      result[level].push(cur.val)
+      cur.left && dfs(cur.left, level + 1)
+      cur.right && dfs(cur.right, level + 1)
+    }
+  }
+  dfs(root, 0)
+  return result
+}