feat: 二叉树问题

Author: sl1673495

二叉树/二叉树的所有路径-257.js

@@ -0,0 +1,18 @@
+let binaryTreePaths = function (root) {
+  let res = []
+  let dfs = (node, path = "") => {
+    if (!node) {
+      return
+    }
+
+    let newPath = path ? `${path}->${node.val}` : `${node.val}`
+    if (!node.left && !node.right) {
+      res.push(newPath)
+    }
+
+    dfs(node.left, newPath)
+    dfs(node.right, newPath)
+  }
+  dfs(root)
+  return res
+}

二叉树/左叶子之和-404.js

@@ -0,0 +1,22 @@
+let sumOfLeftLeaves = function (root) {
+  let sum = 0
+
+  let dfs = (node) => {
+    if (!node) return
+
+    if (isLeaf(node.left)) {
+      sum += node.left.val
+    }
+
+    dfs(node.left)
+    dfs(node.right)
+  }
+
+  dfs(root)
+
+  return sum
+}
+
+function isLeaf(node) {
+  return !!node && !node.left && !node.right
+}

二叉树/路径总和-112.js

@@ -0,0 +1,14 @@
+let hasPathSum = function (root, sum) {
+  if (!root) {
+    return false
+  }
+  // 叶子节点 判断当前的值是否等于 sum 即可
+  if (!root.left && !root.right) {
+    return root.val === sum
+  }
+
+  return (
+    hasPathSum(root.left, sum - root.val) ||
+    hasPathSum(root.right, sum - root.val)
+  )
+}