一棵二叉樹原本是搜索二叉樹，但是其中有兩個節點調換了位置，使得這棵二叉樹不再是搜索二叉樹，請找到這兩個錯誤節點并返回。已知二叉樹中所有節點的 值都不一樣，給定二叉樹的頭節點 head，返回一個長度為 2 的二叉樹節點類型 的數組 errs，errs[0]表示一個錯誤節點，errs[1]表示另一個錯誤節點。

如果在原問題中得到了這兩個錯誤節點，我們當然可以通過交換兩個節點的節點 值的方式讓整棵二叉樹重新成為搜索二叉樹。但現在要求你不能這么做，而是在 結構上完全交換兩個節點的位置，請實現調整的函數。

public static class Node{public int value;public Node left;public Node right;public Node(int data){this.value = data;} }public static Node[] getTwoErrNodes(Node head){Node[] errs = new Node[2];if(head == null){return errs;}Stack<Node> stack = new Stack<Node>();Node pre = null;while(!stack.isEmpty() || head != null){if(head != null){stack.push(head);head = head.left;}else{head = stack.pop();if(pre != null && pre.value > head.value){errs[0] = errs[0] == null ? pre : errs[0];errs[1] = head;}pre = head;head = haed.right;}}return errs; }public static Node[] getTwoErrParents(Node head, Node e1, Node e2){Node[] parents = new Node[2];if(head == null){return parents;}Stack<Node> stack = new Stack<Node>();while(!stack.isEmpty() || head != null){if(head != null){stack.push(head);head = head.left;}else{head = stack.pop();if(head.left == e1 || head.right == e1){parents[0] = head;}if(head.left == e2 || head.right == e2){parents[1] = head;}head = head.right;}}return parents; }public static Node recoverTree(Node head){Node[] errs = getTwoErrNodes(head);Node[] parents = getTwoErrParents(head, err[0], err[1]);Node e1 = errs[0];Node e1P = parents[0];Node e1L = e1.left;Node e1R = e1.right;Node e2 = errs[1];Node e2P = parents[1];Node e2L = e2.left;Node e2R = e2.right;if(e1 == head){if(e1 == e2P){e1.left = e2L;e1.right = e2R;e2.right = e1;e2.left = e1L;}else if(e2P.left == e2){e2P.left = e1;e2.left = e1L;e2.right = e1R;e1.left = e2L;e1.right = e2R;}else{e2P.right = e1;e2.left = e1L;e2.right = e1R;e1.left = e2L;e1.right = e2R;}head = e2;}else if(e2 == head){if(e2 == e1P){e2.left = e1L;e2.right = e1R;e1.left = e2;e1.right = e2R;}else if(e1P.left = e1){e1P.left = e2;e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1R;}else{e1P.right = e2;e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1R;}head = e1;}else{if(e1 == e2P){if(e1P.left == e1){e1P.left = e2;e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1;}else{e1P.right = e2;e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1;}}else if(e2 == e1P){if(e2P.left == e2){e2P.left = e1;e2.left = e1L;e2.right = e1R;e1.left = e2;e1.right = e2R;}else{e2P.right = e1;e2.left = e1L;e2.right = e1R;e1.left = e2;e1.right = e2R;}}else{if(e1P.left == e1){if(e2P.left == e2){e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1R;e1P.left = e2;e2P.left = e1;}else{e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1R:e1P.left = e2;e2P.right = e1;}}else{if(e2P.left == e2){e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1R；e1P.right = e2;e2P.left = e1;}else{e1.left = e2L;e1.right = e2R;e2.left = e1L;e2.right = e1R;e1P.right = e2;e2P.right = e1;}}}}return head; }public static void printTree(Node head){System.out.println("Binary Tree:");printInOrder(head, 0, "H", 17);System.out.println()； }

總結

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