新增一棵二叉树，求最大通路长度（即最大左右子树高度之和）答案

Author: SJshenjian

06.头条篇/一棵二叉树，求最大通路长度（即最大左右子树高度之和）.md

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+#### **题目**：一棵二叉树，求最大通路长度（即最大左右子树高度之和）  
+
+#### **参考答案**：  
+
+该题与leetcode第104题同题型，定义TreeNode结构如下：  
+
+```java
+class TreeNode {
+
+    int val;
+    TreeNode left;
+    TreeNode right;
+
+    public TreeNode(int val) {
+        this.val = val;
+    }
+}
+```
+
+解法一(递归求解)   
+```java
+class Solution {
+
+    public int maxHeight(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        return maxChildHeight(root.left) + maxChildHeight(root.right);
+    }
+
+    public int maxChildHeight(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        int leftHeight = maxChildHeight(root.left);
+        int rightHeight = maxChildHeight(root.right);
+        return Math.max(leftHeight, rightHeight) + 1;
+    }
+}
+```
+
+解法二(迭代求解)
+```java
+public class Solution {
+
+    public int maxHeight(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        return maxChildHeight(root.left) + maxChildHeight(root.right);
+    }
+
+    public int maxChildHeight(TreeNode root) {
+        int height = 0;
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.add(root);
+
+        while (!queue.isEmpty()) {
+            int size = queue.size();
+            for (int i = 0; i < size; i++) {
+                TreeNode node = queue.poll();
+                height++;
+                if (node.left != null) {
+                    queue.add(node.left);
+                }
+                if (node.right != null) {
+                    queue.add(node.right);
+                }
+            }
+        }
+        return height;
+    }
+}
+```