Merge pull request #717 from ironartisan/master

修正递归算法的时间与空间复杂度分析.md输入参数

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -607,6 +607,7 @@ class Solution {
         for (int i = inLeft; i < inRight; i++) {
             if (inorder[i] == rootVal) {
                 rootIndex = i;
+                break;
             }
         }
         // 根据rootIndex划分左右子树

problems/0112.路径总和.md

@@ -411,6 +411,31 @@ class solution {
 }
 ```
 
+```java
+// 解法2
+class Solution {
+    List<List<Integer>> result;
+    LinkedList<Integer> path;
+    public List<List<Integer>> pathSum (TreeNode root,int targetSum) {
+        result = new LinkedList<>();
+        path = new LinkedList<>();
+        travesal(root, targetSum);
+        return result;
+    }
+    private void travesal(TreeNode root,  int count) {
+        if (root == null) return;
+        path.offer(root.val);
+        count -= root.val;
+        if (root.left == null && root.right == null && count == 0) {
+            result.add(new LinkedList<>(path));
+        }
+        travesal(root.left, count);
+        travesal(root.right, count);
+        path.removeLast(); // 回溯
+    }
+}
+```
+
 ## python
 
 0112.路径总和

problems/0257.二叉树的所有路径.md

@@ -367,6 +367,43 @@ class Solution {
     }
 }
 ```
+```java
+// 解法2
+class Solution {
+    /**
+     * 迭代法
+     */
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> result = new ArrayList<>();
+        if (root == null)
+            return result;
+        Stack<Object> stack = new Stack<>();
+        // 节点和路径同时入栈
+        stack.push(root);
+        stack.push(root.val + "");
+        while (!stack.isEmpty()) {
+            // 节点和路径同时出栈
+            String path = (String) stack.pop();
+            TreeNode node = (TreeNode) stack.pop();
+            // 若找到叶子节点
+            if (node.left == null && node.right == null) {
+                result.add(path);
+            }
+            //右子节点不为空
+            if (node.right != null) {
+                stack.push(node.right);
+                stack.push(path + "->" + node.right.val);
+            }
+            //左子节点不为空
+            if (node.left != null) {
+                stack.push(node.left);
+                stack.push(path + "->" + node.left.val);
+            }
+        }
+        return result;
+    }
+}
+```
 
 Python：
 ```Python

problems/0617.合并二叉树.md

@@ -274,7 +274,7 @@ class Solution {
 
 ```Java
 class Solution {
-   // 迭代
+    // 使用栈迭代
     public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
         if (root1 == null) {
             return root2;
@@ -310,6 +310,43 @@ class Solution {
     }
 }
 ```
+```java
+class Solution {
+    // 使用队列迭代
+    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
+        if (root1 == null) return root2;
+        if (root2 ==null) return root1;
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.offer(root1);
+        queue.offer(root2);
+        while (!queue.isEmpty()) {
+            TreeNode node1 = queue.poll();
+            TreeNode node2 = queue.poll();
+            // 此时两个节点一定不为空，val相加
+            node1.val = node1.val + node2.val;
+            // 如果两棵树左节点都不为空，加入队列
+            if (node1.left != null && node2.left != null) {
+                queue.offer(node1.left);
+                queue.offer(node2.left);
+            }
+            // 如果两棵树右节点都不为空，加入队列
+            if (node1.right != null && node2.right != null) {
+                queue.offer(node1.right);
+                queue.offer(node2.right);
+            }
+            // 若node1的左节点为空，直接赋值
+            if (node1.left == null && node2.left != null) {
+                node1.left = node2.left;
+            }
+            // 若node2的左节点为空，直接赋值
+            if (node1.right == null && node2.right != null) {
+                node1.right = node2.right;
+            }
+        }
+        return root1;
+    }
+}
+```
 
 ## Python 
 

problems/前序/递归算法的时间与空间复杂度分析.md

@@ -166,7 +166,7 @@ void time_consumption() {
             system_clock::now().time_since_epoch()
         );
 
-        fibonacci_3(0, 1, n);
+        fibonacci_3(1, 1, n);
 
         milliseconds end_time = duration_cast<milliseconds >(
             system_clock::now().time_since_epoch()