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                二叉樹: ![](https://pic4.zhimg.com/80/v2-0fcc909a2787cd92c1315481748d2b57_720w.jpg) 前序遍歷A-B-D-F-G-H-I-E-C 中序遍歷F-D-H-G-I-B-E-A-C 后序遍歷F-H-I-G-D-E-B-C-A **前序(根左右),中序(左根右),后序(左右根)** 例題1: 已知某二叉樹的前序遍歷為A-B-D-F-G-H-I-E-C,中序遍歷為F-D-H-G-I-B-E-A-C,請還原這顆二叉樹。 解題思路: 從前序遍歷中,我們確定了根結點為A,在從中序遍歷中得出 F-D-H-G-I-B-E在根結點的左邊,C在根結點的右邊,那么我們就可以構建我們的二叉樹的雛形。 ![](https://pic1.zhimg.com/80/v2-117aade0ca15b451e9a542b6852631f0_720w.png) 那么剩下的前序遍歷為B-D-F-G-H-I-E,中序遍歷為F-D-H-G-I-B-E, B就是我們新的“根結點”,從中序遍歷中得出F-D-H-G-I在B的左邊,E在B的右邊,繼續構建 ![](https://pic1.zhimg.com/80/v2-6b585a528009e0978254153dbb43ef70_720w.jpg) 那么剩下的前序遍歷為D-F-G-H-I,中序遍歷為F-D-H-G-I,D就是我們新的“根結點”,從中序遍歷中得出F在D的左邊,H-G-I在D的右邊,繼續構建 ![](https://pic4.zhimg.com/80/v2-99457807b6951cd6a63a46c7a48ef0bb_720w.jpg) 那么剩下的前序遍歷為G-H-I,中序遍歷為H-G-I,G就是我們新的“根結點”,從中序遍歷中得出H在G的左邊,I在G的右邊,繼續構建 ![](https://pic1.zhimg.com/80/v2-85c57eed2a58fd0bddcfd878ed2f13f4_720w.jpg) 例題2: 已知某二叉樹的中序遍歷為F-D-H-G-I-B-E-A-C,后序遍歷為F-H-I-G-D-E-B-C-A,請還原這顆二叉樹。 解題思路: 從后序遍歷中,我們確定了根結點為A,在從中序遍歷中得出 F-D-H-G-I-B-E 在根結點的左邊,C在根結點的右邊,那么我們就可以構建我們的二叉樹的雛形。之后就是新根節點B,FDHGI在根左,E在根右。在之后就是新根D,F根左,HGI根右,然后就差不多了。 和前序和中序還原二叉樹一樣,我們同理可以通過中序和后序還原二叉樹。 ![](https://pic2.zhimg.com/80/v2-d7e98be21ee70faadaaf73a3c697e6c9_720w.jpg) **光有前序遍歷和后序遍歷是無法還原二叉樹的。**
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