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Merge pull request #1397 from wzqwtt/tree09
添加(0404.左叶子之和、0513.找树左下角的值)Scala版本
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@ -516,6 +516,44 @@ int sumOfLeftLeaves(struct TreeNode* root){
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}
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```
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## Scala
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**递归:**
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```scala
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object Solution {
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def sumOfLeftLeaves(root: TreeNode): Int = {
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if(root == null) return 0
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var midValue = 0
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if(root.left != null && root.left.left == null && root.left.right == null){
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midValue = root.left.value
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}
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// return关键字可以省略
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midValue + sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right)
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}
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}
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```
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**迭代:**
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```scala
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object Solution {
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import scala.collection.mutable
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def sumOfLeftLeaves(root: TreeNode): Int = {
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val stack = mutable.Stack[TreeNode]()
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if (root == null) return 0
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stack.push(root)
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var sum = 0
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while (!stack.isEmpty) {
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val curNode = stack.pop()
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if (curNode.left != null && curNode.left.left == null && curNode.left.right == null) {
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sum += curNode.left.value // 如果满足条件就累加
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}
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if (curNode.right != null) stack.push(curNode.right)
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if (curNode.left != null) stack.push(curNode.left)
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}
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sum
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}
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -546,7 +546,50 @@ func findBottomLeftValue(_ root: TreeNode?) -> Int {
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}
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```
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## Scala
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递归版本:
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```scala
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object Solution {
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def findBottomLeftValue(root: TreeNode): Int = {
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var maxLeftValue = 0
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var maxLen = Int.MinValue
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// 递归方法
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def traversal(node: TreeNode, leftLen: Int): Unit = {
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// 如果左右都为空并且当前深度大于最大深度,记录最左节点的值
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if (node.left == null && node.right == null && leftLen > maxLen) {
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maxLen = leftLen
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maxLeftValue = node.value
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}
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if (node.left != null) traversal(node.left, leftLen + 1)
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if (node.right != null) traversal(node.right, leftLen + 1)
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}
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traversal(root, 0) // 调用方法
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maxLeftValue // return关键字可以省略
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}
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}
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```
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层序遍历:
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```scala
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import scala.collection.mutable
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def findBottomLeftValue(root: TreeNode): Int = {
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val queue = mutable.Queue[TreeNode]()
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queue.enqueue(root)
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var res = 0 // 记录每层最左侧结果
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while (!queue.isEmpty) {
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val len = queue.size
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for (i <- 0 until len) {
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val curNode = queue.dequeue()
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if (i == 0) res = curNode.value // 记录最最左侧的节点
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if (curNode.left != null) queue.enqueue(curNode.left)
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if (curNode.right != null) queue.enqueue(curNode.right)
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}
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}
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res // 最终返回结果,return关键字可以省略
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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