From db3b8fffd7cd0dce41962c3a2d55e313c13d3a85 Mon Sep 17 00:00:00 2001
From: liu <3330176274@qq.com>
Date: Sun, 16 Feb 2025 22:40:18 +0800
Subject: [PATCH 1/3] =?UTF-8?q?0654=E6=9C=80=E5=A4=A7=E4=BA=8C=E5=8F=89?=
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diff --git "a/pics/654\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.png" "b/pics/654\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.png"
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diff --git "a/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md" "b/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
index fed9b2b991..7dd3f6f669 100644
--- "a/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
+++ "b/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
@@ -212,6 +212,96 @@ root->right = traversal(nums, maxValueIndex + 1, right);
第二版相应的终止条件，是遇到空节点，也就是数组区间为0，就终止了。
+### 非递归方法
+
+将原数组转化为二叉树时，使数字飘起来，可以想象：越大的数字飘得越高（越接近根节点）。
+
+如原题 【3,2,1,6,0,5】：
+
+<img src="F:\新建文件夹\leetcode-master\pics\654最大二叉树.png" alt="image-20250216223034714" style="zoom:67%;" />
+
+以 **从左到右遍历数组** 的方式构造最大二叉树，避免了递归方法带来的额外栈空间消耗，时间效能上也优一些。
+
+1. 首先创建一个存放各个节点的地址的vector， 可以先把各个节点创建出来，也可以在遍历的过程中创建。
+
+2. 遍历 `nums` 逐步构造树
+
+```cpp
+for (int i = 1; i < nums.size(); i++) {
+ TreeNode* t = new TreeNode(nums[i]);
+ arr.push_back(t);
+```
+
+- 依次读取 `nums[i]` 并新建 `TreeNode`，同时存入 `arr` 数组。
+
+3. **找到 `nums[i]` 前面第一个比它大的数`nump[j]`，则 i 对应的节点应该是 j 节点的右子树，并把 j 原来的右子树(在数组的位置是 i 左边)放在 i 的左子树上。**
+
+```cpp
+int j = i - 1;
+for (; j >= 0; j--) {
+ if (nums[j] > nums[i]) {
+ TreeNode* l = arr[j]->right;
+ arr[j]->right = t;
+ t->left = l;
+ break;
+ }
+}
+```
+
+- 从 `nums[i]` **向左遍历**，找到 **第一个比 `nums[i]` 大的数 `nums[j]`**。
+- 构造父子关系：
+ - `nums[i]` 作为 `nums[j]` 的**右子节点**。
+ - 若 `nums[j]` 原本已有右子树 `l`，则 `l` 作为 `nums[i]` 的**左子节点**。
+
+4. .处理 `nums[i]` 是当前最大值的情况
+
+```
+cpp复制编辑if (j == -1) {
+ t->left = root;
+ root = t;
+}
+```
+
+- 如果 `nums[i]` 比前面所有数都大，则为新的根节点：
+ - 设 `t->left = root`，即将 `root` 作为 `t` 的左子树。
+ - 更新 `root = t`，因为 `nums[i]` 是当前最大值，必须作为新的根节点。
+
+代码如下:
+
+```cpp
+class Solution {
+ public:
+ TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
+ // new一个根节点
+ TreeNode* root = new TreeNode(nums[0]);
+ // 创建一个存放各个node地址的数组，方便修改，与nums顺序一致
+ vector<TreeNode*> arr;
+ // 先把第一个节点设为根节点，放入数组
+ arr.push_back(root);
+ for(int i = 1; i < nums.size(); i++) {
+ TreeNode* t = new TreeNode(nums[i]);
+ arr.push_back(t);
+ int j = i-1;
+ for(; j >= 0; j--) {
+ /* 找到前面第一个大于该数的已有节点 */
+ if(nums[j] > nums[i]){
+ TreeNode* l = arr[j]->right;
+ arr[j]->right = t;
+ t->left = l;
+ break;
+ }
+ }
+ if(j == -1){
+ /* 若没有更大的，说明这个数是目前最大的，设为根节点 */
+ t->left = root;
+ root = t;
+ }
+ }
+ return root;
+ }
+ };
+```
+
## 总结
From c531422db92a0d09d8b9ca6f626b172c7aba740e Mon Sep 17 00:00:00 2001
From: liu <3330176274@qq.com>
Date: Sun, 16 Feb 2025 22:46:02 +0800
Subject: [PATCH 2/3] =?UTF-8?q?0654=E6=9C=80=E5=A4=A7=E4=BA=8C=E5=8F=89?=
=?UTF-8?q?=E6=A0=91.md=E5=A2=9E=E5=8A=A0=E6=96=B0=E9=9D=9E=E9=80=92?=
=?UTF-8?q?=E5=BD=92=E7=AE=97=E6=B3=95=EF=BC=8C=E4=B8=80=E5=BC=A0=E5=9B=BE?=
=?UTF-8?q?=E7=89=87=E6=94=BE=E5=9C=A8=E4=BA=86pics=E6=96=87=E4=BB=B6?=
=?UTF-8?q?=E5=A4=B9=E4=B8=AD?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
...34\200\345\244\247\344\272\214\345\217\211\346\240\221.md" | 4 ++--
1 file changed, 2 insertions(+), 2 deletions(-)
diff --git "a/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md" "b/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
index 7dd3f6f669..18f9162ed9 100644
--- "a/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
+++ "b/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
@@ -255,8 +255,8 @@ for (; j >= 0; j--) {
4. .处理 `nums[i]` 是当前最大值的情况
-```
-cpp复制编辑if (j == -1) {
+```cpp
+if (j == -1) {
t->left = root;
root = t;
}
From d89a14b6435029cf09bce96dcd376755003313f7 Mon Sep 17 00:00:00 2001
From: liu <3330176274@qq.com>
Date: Sun, 16 Feb 2025 22:47:39 +0800
Subject: [PATCH 3/3] =?UTF-8?q?0654=E6=9C=80=E5=A4=A7=E4=BA=8C=E5=8F=89?=
=?UTF-8?q?=E6=A0=91.md=E5=A2=9E=E5=8A=A0=E6=96=B0=E9=9D=9E=E9=80=92?=
=?UTF-8?q?=E5=BD=92=E7=AE=97=E6=B3=95=EF=BC=8C=E4=B8=80=E5=BC=A0=E5=9B=BE?=
=?UTF-8?q?=E7=89=87=E6=94=BE=E5=9C=A8=E4=BA=86pics=E6=96=87=E4=BB=B6?=
=?UTF-8?q?=E5=A4=B9=E4=B8=AD?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
...\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md" | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git "a/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md" "b/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
index 18f9162ed9..f45adb0a4d 100644
--- "a/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
+++ "b/problems/0654.\346\234\200\345\244\247\344\272\214\345\217\211\346\240\221.md"
@@ -218,7 +218,7 @@ root->right = traversal(nums, maxValueIndex + 1, right);
如原题 【3,2,1,6,0,5】：
-<img src="F:\新建文件夹\leetcode-master\pics\654最大二叉树.png" alt="image-20250216223034714" style="zoom:67%;" />
+<img src="..\pics\654最大二叉树.png" alt="image-20250216223034714" style="zoom:67%;" />
以 **从左到右遍历数组** 的方式构造最大二叉树，避免了递归方法带来的额外栈空间消耗，时间效能上也优一些。