Java实现二叉树的相关操作

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// 求二叉树的深度    public static int BiTreeDepth(BitNode T) {        int depthval, depthLeft, depthRight;        if (T == null)            depthval = 0;        else if (T.lchild == null && T.rchild == null)            depthval = 1;        else {            depthLeft = BiTreeDepth(T.lchild);            depthRight = BiTreeDepth(T.rchild);            depthval = 1 + (depthLeft > depthRight ? depthLeft : depthRight);        }        return depthval;    }// 求data所对应结点的层数，如果对象不在树中,结果返回-1;否则结果返回该对象在树中所处的层次,规定根节点为第一层    public static int level(BitNode bitNode, int data) {        int leftLevel, rightLevel;        if (bitNode == null)            return -1;        if (data == bitNode.data)            return 1;        leftLevel = bitNode.lchild == null ? -1 : level(bitNode.lchild, data);        rightLevel = bitNode.rchild == null ? -1 : level(bitNode.rchild, data);        if (leftLevel < 0 && rightLevel < 0)            return -1;        return leftLevel > rightLevel ? leftLevel + 1 : rightLevel + 1;    }    // 求二叉树叶子节点的总数    public static int leafNum(BitNode tree) {        if (tree == null)            return 0;        else {            int left = leafNum(tree.lchild);            int right = leafNum(tree.rchild);            if (tree.lchild == null && tree.rchild == null)                return left + right + 1;            else                return left + right;        }    }    // 求二叉树父节点个数    public static int fatherNodes(BitNode tree) {        if (tree == null || (tree.lchild == null && tree.rchild == null))            return 0;        else {            int left = fatherNodes(tree.lchild);            int right = fatherNodes(tree.rchild);            return left + right + 1;        }    }    // 求只有一个孩子结点的父节点个数    public static int oneChildFather(BitNode tree) {        int left, right;        if (tree == null || (tree.rchild == null && tree.lchild == null))            return 0;        else {            left = oneChildFather(tree.lchild);            right = oneChildFather(tree.rchild);            if ((tree.lchild != null && tree.rchild == null)                    || (tree.lchild == null && tree.rchild != null))                return left + right + 1;            else                return left + right;/* 加1是因为要算上根节点 */        }    }    // 求二叉树只拥有左孩子的父节点总数    public static int leftChildFather(BitNode tree) {        if (tree == null)            return 0;        else {            int left = leftChildFather(tree.lchild);            int right = leftChildFather(tree.rchild);            if ((tree.lchild != null && tree.rchild == null))                return left + right + 1;            else                return left + right;        }    }    // 求二叉树只拥有左孩子的父节点总数    public static int rightChildFather(BitNode tree) {        if (tree == null)            return 0;        else {            int left = leftChildFather(tree.lchild);            int right = leftChildFather(tree.rchild);            if ((tree.lchild == null && tree.rchild != null))                return left + right + 1;            else                return left + right;        }    }    // 计算有两个节点的父节点的个数    public static int doubleChildFather(BitNode tree) {        int left, right;        if (tree == null)            return 0;        else {            left = doubleChildFather(tree.lchild);            right = doubleChildFather(tree.rchild);            if (tree.lchild != null && tree.rchild != null)                return (left + right + 1);/* 加1是因为要算上根节点 */            else                return (left + right);        }    }    // 将树中的每个节点的孩子对换位置    public static void exChange(BitNode tree) {        if (tree == null)            return;        if (tree.lchild != null)            exChange(tree.lchild);        if (tree.rchild != null)            exChange(tree.rchild);        BitNode temp = tree.lchild;        tree.lchild = tree.rchild;        tree.rchild = temp;    }    // 递归求所有结点的和    public static int getSumByRecursion(BitNode tree) {        if (tree == null) {            return 0;        } else {            int left = getSumByRecursion(tree.lchild);            int right = getSumByRecursion(tree.rchild);            return tree.data + left + right;        }    }
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